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PHP - How To Use Select Box For Displaying A Database Value
I create list box which value contains from the table of database
but now i want to display the data related to the value of list box. please any body can help me to solve the problem My table contains id, category_name, title, discription and list box contain the value from the filed category_name of the table now i want to display title and discription according to the category_name. please send me code i am totally confused.. Similar Tutorialshi i am in situation to display input box along with select option , here is my code <?php $run= $dbc->db("SELECT * FROM `table` WHERE `id`='$id'"); ?> <select name="service"> <?php while ($result = mysql_fetch_array($run)) { ?> <option value="<?php echo $result['id'] ; ?>"><?php echo $result['service'] ; ?></option> <?php // This is problem i need to get this as hidden field but it breaks my select options. ?> <input type="hidden" value="<?php echo $result['data'] ; ?>" name="data" /> <?php } ?> </select> so here i am stuck <input type="hidden" value="<?php echo $result['id'] ; ?>" name="data" /> when i repeat in while loop it breaks options box and display other options without in select box. any help please thanks Below is a page which is supposed to output the name, blog contribution and picture of contributing members of a website. <div id="blog_content" class="" style="height:90%; width:97%; border:5px solid #c0c0c0; background-color: #FFFFFF;"> <!--opens blog content--> <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //include the config file require_once("config.php"); //Define the query. Select all rows from firstname column in members table, title column in blogs table,and entry column in blogs table, sorting in ascneding order by the title entry, knowing that the id column in mebers table is the same as the id column in blogs table. $sql = "SELECT blogs.title,blogs.entry,members.firstname,images.image FROM blogs LEFT JOIN members ON blogs.member_id = members.member_id LEFT JOIN images ON blogs.member_id = images.member_id ORDER BY blogs.title ASC "; $query = mysql_query($sql); if($query !== false && mysql_num_rows($query) > 0) { while(($row = mysql_fetch_assoc($query)) !== false) { echo '<div id="blog_content1" style="float:left; position:relative;bottom:18px;left:13px; background-color: #FFFFFF; height:16.7%; width:100%; border:0px none none;" <!--opens blog_content1 same as main center top 1 and 2 from index page everything scaled down by a factor of 3, heightwise--> <div class="red_bar" style="height:3%; width:100%; border:1px solid #959595;"> <!--a--> <div class="shade1" style="height:5px; width:100%; border:0px none none;"> </div> <div class="shade2" style="height:5px; width:100%; border:0px none none"> </div> <div class="shade3" style="height:5px%; width:100%; border:0px none none"> </div> </div> <!-- closes red bar--> <div class="content" style="height:28.3%; width:100%; border:0px none none;"> <!----> <div class="slideshow" id="keylin" style="float:left; width:20%; border:0px none none;"> <!--a--> <div><img header("Content-type: image/jpeg"); name="" alt="" id="" height="105" width="105" src="$row[image]" /></div> </div> <!-- closes pic--> <div class="content_text" style="float:right; position:relative;top:7px;left:0px; max-height:150px; width:78.5%; border-width:4.5px; border-bottom-style:solid; border-right-style:solid; border-color:#c0c0c0; "> <!--a-->'; echo "<h3>".$row['title']."</h3>"; echo "<p>" .$row['entry']."<br />".$row['firstname']."</p>"; echo '</div> <!-- closes content text--> </div> <!-- closes content--> </div> <!-- closes blog_content1-->'; } } else if($query == false) { echo "<p>Query was not successful because:<strong>".mysql_error()."</strong></p>"; echo "<p>The query being run was \"".$sql."\"</p>"; } else if($query !== false && mysql_num_rows($query) == 0) { echo "<p>The query returned 0 results.</p>"; } mysql_close(); //Close the database connection. ?> </div> <!-- closes blog content--> The select query is designed to retrieve all the blog contributions(represented by the fields blogs.title and blogs.entry) from the database, alongside the contributing member (member.firstname) and the member's picture(images.image), using the member_id column to join the 3 tables involved, and outputs them on the webpage. The title, entry and firstname values are successfully displayed on the resulting page. However, I can't seem to figure out how to get the picture to be displayed. Note that the picture was successfully stored in the database and I was able to view it on a separate page using a simple select query. It is now just a question of how to get it to display on this particularly crowded page. Anyone knows how I can output the picture in the img tag? I tried placing the header("Content-type: image/jpeg"); statement at the top of the php segment, then just right below the select query and finally just right above the img tag, but in every case, I just got a big white blank page starring at me. How and where should I place the header statement? And what else am I to do to get this picture displayed? Any help is appreciated. I want my mysql database to be displayed across the screen similar to the below Entry 1 Entry 2 Entry 3 Entry 4 Entry 5 eventually with a image above the name, but for now just as above.
How would I go about this?
I am trying to display data from a database from a form entry here is the php <?php include('dbconnect.php'); @mysql_select_db($database) or die( "Unable to select database"); $query="SELECT * FROM child_info"; $result=mysql_query($query); $num=mysql_numrows($result); mysql_close(); echo "<b><center>Database Output</center></b><br><br>"; $i=0; while ($i < $num) { $field1-name=mysql_result($result,$i,"file_number"); $field2-name=mysql_result($result,$i,"first_name"); $field3-name=mysql_result($result,$i,"middle_name"); $field4-name=mysql_result($result,$i,"last_name"); $field5-name=mysql_result($result,$i,"birthdate"); $field6-name=mysql_result($result,$i,"gender"); $field7-name=mysql_result($result,$i,"features"); $field8-name=mysql_result($result,$i,"diagnosis"); $field9-name=mysql_result($result,$i,"description"); echo "<b>$field1-name $field2-name2</b><br>$field3-name<br>$field4-name<br>$field5-name<hr><br>"; $i++; } ?> here is the form I am using <form name="child_info" action="selectdata.php" method="post" id="child_info"> <table width="444" align="center" > <tr> <td> Search by Name: </td> <td> First Name:<input type="text" class="form-textbox " id="first_name" name="first_name" size="20" /><br /> Last Name:<input type="text" class="form-textbox " id="last_name" name="last_name" size="20" /> </td> </tr> <tr> <td width="208"> Choose Male or Female: </td> <td width="224"> <input type="radio" name="gender" value="male" /> Male <input type="radio" name="gender" value="Female" /> Female </td> </tr> <tr> <td> Choose age range: </td> <td> <select name="first_age" id="first_age"> <option value="00">From</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> <option value="6">6</option> <option value="7">7</option> <option value="8">8</option> <option value="9">9</option> <option value="10">10</option> <option value="11">11</option> <option value="12">12</option> <option value="13">13</option> <option value="14">14</option> <option value="15">15</option> <option value="16">16</option> <option value="17">17</option> <option value="18">18</option> </select> <select name="second_age" id="second_age"> <option value="00">To</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> <option value="6">6</option> <option value="7">7</option> <option value="8">8</option> <option value="9">9</option> <option value="10">10</option> <option value="11">11</option> <option value="12">12</option> <option value="13">13</option> <option value="14">14</option> <option value="15">15</option> <option value="16">16</option> <option value="17">17</option> <option value="18">18</option> </select> </td> </tr> <tr> <td> </td> <td> </td> </tr> <tr> <td> </td> <td> <div align="right"> <input type="submit" name="submit" id="submit" value="submit" /> <input type="reset" name="reset" id="reset" value="reset" /> </div></td> </tr> </table> </form> first problem is getting the form to use the php second problem is when i try to use the php alone is I get this error Parse error: syntax error, unexpected '=' in /home/fathersh/public_html/selectdata.php on line 17 17 is highlighted above in the php So I've followed this code, corrected about 12 error, talked to my hosting and I am so done. I have tried everything but the error won't go away. The code is pasted below. As it's really late any help would be appreciated that would make my day the next day...
<?PHP So I have an simple account centre up, and i'm wanting to display their 'Name' 'Username' and 'Email' as part of their details. But I have one problem... My code doesn't seem to be getting the data from my database... It may be messy to some people, just warning you! Code: [Select] <?php session_start(); if($_SESSION['username']){ $connect = mysql_connect("****","****","****") or die("Could not connect to database."); mysql_select_db("****") or die ("Could not find database!"); $sql = mysql_query("SELECT * FROM login"); $username = $rows['username']; $email = $rows['email']; $rows = mysql_fetch_assoc($sql); echo "<p>"; } else header("location: suggestion.html"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> <link href="css/style.css" rel="stylesheet" type="text/css" /> <style type="text/css"> body { background-color: #CCC; } body,td,th { color: #000; font-family: "MS Serif", "New York", serif; } </style> </head> <body> <div id="wrap"> <!--Header--> <div id="header_member"> </div> <!--Log out and time--> <div id="info"> <div id="date"><script type="text/javascript"> var currentDate = new Date() var day = currentDate.getDate() var month = currentDate.getMonth() + 1 var year = currentDate.getFullYear() document.write("<b>" + day + "/" + month + "/" + year + "</b>") var currentTime = new Date() var hours = currentTime.getHours() var minutes = currentTime.getMinutes()</script> </div> <div id="time"><script type="text/javascript"> var suffix = "AM"; if (hours >= 12) { suffix = "PM"; hours = hours - 12; } if (hours == 0) { hours = 12; } if (minutes < 10) minutes = "0" + minutes document.write("<b>" + hours + ":" + minutes + " " + suffix + "</b>")</script> </div> </div> <div id="logout"><center><?php echo "<a href='logout.php'>Log out.</a>";?></center></div> <!--Main section which will contain everything else--> <div id="member_main"> <div id="member_right"><center> <p><img src="images/accountinf.png" width="175" height="30" /></p> <p>Name: <?php echo $username;?></p> <p>Email: <?php echo $email;?></p> </center> </div> <div id="member_top"><center><?php echo "Welcome, ".$_SESSION['username'];?></center></div> <div id="member_left" align="center"><img src="images/navigation.png" width="105" height="30" /><img src="images/home_member.png" width="105" height="30" /><img src="images/account.png" width="105" height="30" /></div> </div> <!--Footer--> <div id="footer_member"></div> </div> </body> </html> PHP Code Code: [Select] <?php $username=""; $password=""; $database=""; mysql_connect("","",""); @mysql_select_db($database) or die( "Unable to select database"); $query="SELECT * FROM tablename"; $result=mysql_query($query); $num=mysql_numrows($result); mysql_close(); echo "<b><center>Database Output</center></b><br><br>"; $i=0; while ($i < $num) { $field1-name=mysql_result($result,$i,"id"); $field2-name=mysql_result($result,$i,"Location"); $field3-name=mysql_result($result,$i,"Property type"); $field4-name=mysql_result($result,$i,"Number of bedrooms"); $field5-name=mysql_result($result,$i,"Purchase type"); $field6-name=mysql_result($result,$i,"Price range"); echo "<b>$field1-name $field2-name2</b><br>$field3-name<br>$field4-name<br>$field5-name<hr><br>"; $i++; } ?> HTML code for the form Code: [Select] <table id="tb1"> <tr> <td><p class="LOC">Location:</p></td> <td><div id="LC"> <form action="insert.php" method="post"> <select multiple="multiple" size="5" style="width: 150px;" > <option>Armley</option> <option>Chapel Allerton</option> <option>Harehills</option> <option>Headingley</option> <option>Hyde Park</option> <option>Moortown</option> <option>Roundhay</option> </select> </form> </div> </td> <td><p class="PT">Property type:</p></td> <td><div id="PS"> <form action="insert.php" method="post"> <select name="property type" style="width: 170px;"> <option value="none" selected="selected">Any</option> <option value="Houses">Houses</option> <option value="Flats / Apartments">Flats / Apartments</option> </select> </form> </div> </td><td> <div id="ptype"> <form action="insert.php" method="post"> <input type="radio" class="styled" name="ptype" value="forsale"/> For Sale <p class="increase"> <input type="radio" class="styled" name="ptype" value="forrent"/> To Rent </p> <p class="increase"> <input type="radio" class="styled" name="ptype" value="any"/> Any </p> </form> </div> </td> </tr> </table> <div id="table2"> <table id="NBtable"> <tr> <td><p class="NBS">Number of bedrooms:</p></td> <td><div id="NB"> <form action="insert.php" method="post"> <select name="number of bedrooms"> <option value="none" selected="selected">No Min</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> to <select name="number of bedrooms"> <option value="none" selected="selected">No Max</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> <option value="5">5</option> </select> </form> </div> </td> <td><p class="PR">Price range:</p></td> <td><div id="PR"> <form action="insert.php" method="post"> <select name="price range"> <option value="none" selected="selected">No Min</option> <option value="50,000">50,000</option> <option value="60,000">60,000</option> <option value="70,000">70,000</option> <option value="80,000">80,000</option> <option value="90,000">90,000</option> <option value="100,000">100,000</option> <option value="110,000">110,000</option> <option value="120,000">120,000</option> <option value="130,000">130,000</option> <option value="140,000">140,000</option> <option value="150,000">150,000</option> <option value="160,000">160,000</option> <option value="170,000">170,000</option> <option value="180,000">180,000</option> <option value="190,000">190,000</option> <option value="200,000">200,000</option> <option value="210,000">210,000</option> <option value="220,000">220,000</option> <option value="230,000">230,000</option> <option value="240,000">240,000</option> <option value="250,000">250,000</option> <option value="260,000">260,000</option> <option value="270,000">270,000</option> <option value="280,000">280,000</option> <option value="290,000">290,000</option> <option value="300,000">300,000</option> <option value="310,000">310,000</option> <option value="320,000">320,000</option> <option value="330,000">330,000</option> <option value="340,000">340,000</option> <option value="350,000">350,000</option> </select> to <select name="price range"> <option value="none" selected="selected">No Max</option> <option value="50,000">50,000</option> <option value="60,000">60,000</option> <option value="70,000">70,000</option> <option value="80,000">80,000</option> <option value="90,000">90,000</option> <option value="100,000">100,000</option> <option value="110,000">110,000</option> <option value="120,000">120,000</option> <option value="130,000">130,000</option> <option value="140,000">140,000</option> <option value="150,000">150,000</option> <option value="160,000">160,000</option> <option value="170,000">170,000</option> <option value="180,000">180,000</option> <option value="190,000">190,000</option> <option value="200,000">200,000</option> <option value="210,000">210,000</option> <option value="220,000">220,000</option> <option value="230,000">230,000</option> <option value="240,000">240,000</option> <option value="250,000">250,000</option> <option value="260,000">260,000</option> <option value="270,000">270,000</option> <option value="280,000">280,000</option> <option value="290,000">290,000</option> <option value="300,000">300,000</option> <option value="310,000">310,000</option> <option value="320,000">320,000</option> <option value="330,000">330,000</option> <option value="340,000">340,000</option> <option value="350,000">350,000</option> </select> </form> </div> </td> </tr> </table> </div> <form id="submit" action=""> <input type="submit" value="search" /> </form> Hi, I'm trying to joing two tables together, in the resulting PHP coding (See below) the issue I'm having the coding saying select the colour from the colour table where the model of car is the same as the model in the cars table. For example, if the car model is KA and in the table it is KA show the colours. I know this manual fix does the trick. [ $test = "KA"; $query_cols = "SELECT * FROM colours WHERE colours.model = '$test' "; ] But not the soultion any help would be much appericated thank you. [ include "connections/dbconnect.php"; $manfactures = "Ford"; $car_query = "SELECT * FROM cars WHERE make = '$manfactures'"; $car_result = mysql_query($car_query) or die ("Error in query: $car_query. ".mysql_error()); setlocale(LC_MONETARY, 'en_GB'); $fmt = '%i'; if (mysql_num_rows($car_result) > 0) { while ($car_row = @ mysql_fetch_array($car_result)) { $test = "KA"; $query_cols = "SELECT * FROM colours WHERE colours.model = cars.model"; $cols_result = mysql_query($query_cols) or die ("Error in query: $query_cols. ".mysql_error()); print " <table class='details'> <tr> <td rowspan='2'> <img src=\"". $car_row["image"] ."\" alt='" . $car_row["image_alt"] . "' /> </td> <td colspan='2'> <a href='" . $car_row["what_link"] . "'> " . $car_row["model"]." ".$car_row["model_details"] . " </a> </tr> <tr> <td> <p class='info'> RRP:<br/> What Price:<br/> Our Price:<br/> Savings of:<br/> Delivery Time: </p> </td> <td> <p class='info1'> "; echo money_format($fmt, $car_row["rrp"] ); print "<br/>"; echo money_format($fmt, $car_row["what_price"] ); print "<br/>"; echo money_format($fmt, $car_row["our_price"] ); $savings = $car_row["rrp"] - $car_row["our_price"]; print " <br/> <font color=\"red\">"; echo money_format($fmt, $savings ); print " </font><br/> " . $car_row["delivery_time"] . " </p> </td> </tr> <tr> <td> "; while ($cols_row = @ mysql_fetch_array($cols_result)) { ?> <a href='#' onmouseout='hideTooltip()' onmouseover='showTooltip(event,"<?php print "" . $cols_row["colour"] . ""; ?>");return false'> <?php print " <img src=\"". $cols_row["colour_img"] ."\" alt='" . $cols_row["colour_img_alt"] . "' /> "; } print " </td> </tr> "; } } else { echo "Aids!"; } print "</table>"; ?>] Hi guys, it's me Captain Failure, back with more antics. I can't get this to link to the database and was wondering if anyone could help :/ The main code is the second block below but I have posted some of the config file also. Config.php public $db_host = 'localhost'; public $db_user = 'user'; public $db_pass = 'pass'; public $db_name = 'db'; Main page require_once('config.php'); $c = new Config; $query = "DELETE FROM tsclient WHERE id = '".$uid."';"; $db_connection = mysql_connect($c->db_host,$c->db_user,$c->db_pass); if (!$db_connection) { die($errmsg['ERR_MYSQL']); } $db_selection = mysql_select_db($c->db_name,$db_connection); if (!$db_selected) { die($l->errmsg['ERR_DBFAIL']); } mysql_query($query,$db_selection); mysql_close($db_connection); I keep hitting die($l->errmsg['ERR_DBFAIL']); but can't work out why... The use has All Privileges in that database and the user/pass details must be correct because it is actually connecting and not giving out [ph]die($errmsg['ERR_MYSQL']);[/php] Can anyone help? :/ the script succesfuly insert image to the database bt, i cant be able to display it on my pages, any help i will appreciate
Attached Files
saveimage.php 1.15KB
2 downloads
images_tbl.php 192bytes
3 downloads Hi I have got results being displayed after clicking the search button in a form on my home page but it brings up all the results which is ok but how do I get onlt the results a user searches for for example a location or property type etc as its for a property website The coding is below for the results page Also sorry how do I add a background image to the php page, I tried using css but wouldn't work Code: [Select] <style type="text/css"> body {background-image:url('images/greybgone.png');} </style> <?php mysql_connect ("2up2downhomes.com.mysql", "2up2downhomes_c","mD8GsJKQ") or die (mysql_error()); mysql_select_db ("2up2downhomes_c"); echo $_POST['term']; $sql = mysql_query("select * from properties where typeProperty like '%$term%' or location like '%$term%'"); while ($row = mysql_fetch_array($sql)){ echo 'Type of Property: '.$row['typeProperty']; echo '<br/> Number of Bedrooms: '.$row['bedrooms']; echo '<br/> Number of Bathrooms: '.$row['bathrooms']; echo '<br/> Garden: '.$row['garden']; echo '<br/> Description: '.$row['description']; echo '<br/> Price: '.$row['price']; echo '<br/> Location: '.$row['location']; echo '<br/> Image: '.$row['image']; echo '<br/><br/>'; } ?> this is the line in my script that I have to show the image: Code: [Select] $output .= "<img>{$row['disp_pic']}</img></br>\n"; As you can see I added the image tag, but it wont show the actual image. IE shows it as a small square with another small square picture icon in the middle of it (i'm sure you guys know what i mean). Hi I have a text area, that I want to display info pulled from a database. I can get the data to show, But can get each entry of the table to display on it's own line. Example: ob1ob2ob3ob4ob5 Should be: ob1 ob2 ob3 ob4 ob5 CODE: Code: [Select] <textarea id="interest" onfocus="clearInterest()" class="textareacss" style="height:212px;overflow:auto;"><?PHP $newInterestSub = Admin_interests_sub::find_by_cat_id($id); foreach($newInterestSub as $newInterestSubs){ echo $newInterestSubs->interest_sub.'\n'; } ?></textarea> Any help would be great. ok so im sure this is only a small problem but still here it is: im making a shopping list app where users can create a list...when they view the list they can populate it with categories such as frozen food, fruit, veg etc etc...they can then populate categories with items such as apples, potatoes or ice cream etc etc. now i have some data in the database already...and i wanted to display it on the page like this. ASDA SHOPPING LIST fruit apples bananas plums veg potatoes carrots frozen burgers chips ice cream however at the moment with my code it displays like this: ASDA SHOPPING LIST fruit apples bananas plums potatoes carrots burgers chips ice cream veg frozen here is my code: include_once("config_class.php"); $db = new db(); // open up the database object $db->connect(); // connect to the database //getting id of the data from url $id = $_GET['id']; $sql=mysql_query("SELECT listname FROM list WHERE listid=$id") or die("cannot select: ".mysql_error()); $sql2=mysql_query("SELECT catid, category FROM cat WHERE listid=$id") or die("cannot select: ".mysql_error()); $sql3=mysql_query("SELECT items.itemname, items.itemid, cat.catid FROM items, cat WHERE cat.catid=items.catid") or die("cannot select: ".mysql_error()); $temp_cat = ""; $res=mysql_fetch_array($sql); echo "<b>" . $res['listname'] . "</b>" . "<br><br>"; echo "<form action='addcat.php?id=$id' method='post'>"; echo "<input type='text' id='addcat' name='addcat'>"; echo "<input type='submit' value='Add Category'>"; echo "</form>"; while($res2=mysql_fetch_array($sql2)) { echo "<table cellpadding='2' cellspacing='2' width='800'>"; echo "<tr>"; if($res2['category'] != $temp_cat ) { echo "<td width='20%'>"; echo "<b>" . $res2['category'] . "</b>" . "</td>"; echo "<td width='20%'><a href='delcat.php?id=$res2[catid]&id2=$id'>Delete Category</a></td>"; echo "<form action='additem.php?id=$res2[catid]&id2=$id' method='post' name='form1'>"; echo "<td width='20%'>"; echo "<input type='text' name='itemname'></td>"; echo "<td width='20%'>"; echo "<input type='submit' name='Submit' value='Add Item'></td>"; echo "</form>"; echo "</tr>"; $temp_cat=$res2['category']; } while($res3=mysql_fetch_array($sql3)) { echo "<tr>"; echo "<td width='20%'>"; echo "$res3[itemname]" . "</td>"; echo "<td width='20%'>"; echo "<a href='delitem.php'>Delete Item</a>" . "</td>"; echo "</tr>"; } echo "</table>"; } could someone please help me display this correctly? thanks in advance Hi, wondering if somebody can tell me where I'm going wrong (I'm new to all of this). I have the following php code which uploads an image file into my database: Code: [Select] //Connect to database include 'Resources/Include/db.inc.php'; $tmp=$_FILES['image']['tmp_name']; //get users IP $ip=$_SERVER['REMOTE_ADDR']; //Don't do anything if file wasn't selected if (!empty($tmp)) { //Copy file to temporary folder copy($tmp, "./temporary/".$ip.""); //open the copied image, ready to encode into text to go into the database $filename1 = "./temporary/".$ip; $fp1 = fopen($filename1, "rb"); //record the image contents into a variable $contents1 = fread($fp1, filesize($filename1)); $contents1 = addslashes($contents1); //close the file fclose($fp1); $ftype = $_FILES['image']['type']; //insert information into the database if(!mysql_query("INSERT INTO LetterImages (Data,Type,LetterID,Page)"." VALUES ( '$contents1', '$ftype',1,1)")){ echo mysql_error(); } //delete the temporary file we made unlink($filename1); } This seems to work ok, as when I go to the LetterImages table there is now an additional row with a file in the blob field. I then have the following code which is supposed to display the image: Code: [Select] $result=mysql_query("SELECT * FROM LetterImages WHERE LetterID=1 AND Page=1"); //fetch data from database $sqldata=mysql_fetch_array($result); $encoded=stripslashes($sqldata['Data']); $ftype=$sqldata['Type']; //tell the browser what type of image to display header("Content-type: $ftype"); //decode and echo the image data echo $encoded; Instead of displaying an image, however, this just displays pages and pages of incomprehensible data. Can anybody tell me where I'm going horribly wrong? Hey, I have written a script for a very simple PHP wall and comment system. This works fine but the problem I have is displaying the comments. It seems to display the comments associated with the post as well as the comments on the posts above it. I have checked the database and the post ID's are correct. Here is my code: Code: [Select] <?php $wallDisplay = ''; $commentDisplay = ''; $wallDisplaySql = mysql_query("SELECT * FROM wall WHERE to_id='$id' ORDER BY datetime DESC") or die (mysql_error()); while($row = mysql_fetch_array($wallDisplaySql)){ $wallPostId = $row["id"]; $to_id = $row["to_id"]; $from_id = $row["from_id"]; $message = $row["message"]; $dateTime = $row["datetime"]; $getFromData = mysql_query("SELECT username FROM members WHERE id='$from_id'") or die (mysql_error()); while($row2 = mysql_fetch_array($getFromData)){ $wallUsername = $row2['username']; } $displayComments = mysql_query("SELECT * FROM wallComments WHERE wallPostId='$wallPostId' ORDER BY datetime DESC"); while($row3 = mysql_fetch_array($displayComments)){ $wallComment = $row3['comment']; $commentFrom = $row3['from_id']; $commentDate = $row3['datetime']; $getUsername = mysql_query("SELECT username FROM members WHERE id='$commentFrom'"); while($row4 = mysql_fetch_array($getUsername)){ $commentUsername = $row4['username']; } $cheersCheck_pic = "members/$commentFrom/pic1.jpg"; $cheersDefault_pic = "members/0/defaultMemberPic.jpg"; if (file_exists($cheersCheck_pic)) { $cheers_pic = "<img src=\"$cheersCheck_pic?$cacheBuster\" width=\"40px\" />"; } else { $cheers_pic = "<img src=\"$cheersDefault_pic\" width=\"40px\" />"; } $commentDisplay .= '<table width="500px" align="right" cellpadding="4" bgcolor="#FFF"> <tr> <td width="10%" bgcolor="#FFFFFF"><a href="member_profile.php?id=' . $commentFrom . '">' . $cheers_pic . '</a><br /> </td> <td width="90%" bgcolor="#DBE4FD"><a href="member_profile.php?id=' . $commentFrom . '"><span class="blackText">' . $commentUsername . '</span></a> • <span class="blackTetx">' . $commentDate . '<br /><font size="1"></font></span><br /> <span class="blackText">' . $wallComment . '</span></td> </tr> </table>'; } $cheersCheck_pic = "members/$from_id/pic1.jpg"; $cheersDefault_pic = "members/0/defaultMemberPic.jpg"; if (file_exists($cheersCheck_pic)) { $cheers_pic = "<img src=\"$cheersCheck_pic?$cacheBuster\" width=\"40px\" />"; } else { $cheers_pic = "<img src=\"$cheersDefault_pic\" width=\"40px\" />"; } $wallDisplay .= '<table width="100%" align="center" cellpadding="4" bgcolor="#FFF"> <tr> <td width="7%" bgcolor="#FFFFFF"><a href="member_profile.php?id=' . $from_id . '">' . $cheers_pic . '</a><br /> </td> <td width="93%" bgcolor="#DBE4FD"><a href="member_profile.php?id=' . $from_id . '"><span class="blackText">' . $wallUsername . '</span></a> • <span class="blackTetx">' . $dateTime . '<br /><font size="1"></font></span><br /> <span class="blackText">' . $message . '</span></td> </tr> </table> <div id="commentList">' . $commentDisplay . '</div> <div id="comment" align="right"> <form id="comment" name="comment" method="post" action="member_profile.php?id=' .$id. '"> <textarea name="comment" id="comment" rows="1" cols="35"></textarea> <input type="hidden" name="wallPostId" id="wallPostId" value="'. $wallPostId .'" /> <input type="hidden" name="commentFrom" id="commentFrom" value="'. $_SESSION['id'] .'" /> <input type="submit" name="submitComment" id="submitComment" /> </form> </div><br /> '; } ?> I have been looking at it for ages but can think why this is happening. Thanks in advance for any help Hi im not sure if this can be done or not but im trying to do a site without using mysql and i want to be able to compare 3 values and depending on the values have them aranged lowest to highest... for example: Apple = 8 Pear = 3 Bannana = 5 so the results would be displayed like... Pear with a total of 3 bannana with a total of 5 Apple with a total of 8 Is this possible using just PHP or will i need to use Mysql as well... Thank you Chris Hi guys, is it possible to update a field ina a database using a select box. So far i have populated the select box with database values and it works fine, however i dont know how to update the field properly in a database. have a look at the code and tell me what you think. Code: [Select] <?php session_start(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>stock manager</title> </head> <body> <center> <table> <td> <table> <td> <form action='stockview.php' method='POST'> Please Enter a Stock Name and Stock Value <table> <tr> <td> Stock Name: </td> <td> <input name="stockname" type="text" /><BR /> </td> </tr> <tr> <td> Stock Qty: </td> <td> <input name="stockqty" type="text" /> </td> </tr> </table> <input name="submit1" type="submit" value="Add New Stock Items" /> </form> </td> </table> </td> <td> <table> <td> <form action='stockview.php' method='POST' enctype="multipart/form-data"> Please Select from the list the item you wish to update <table> <tr> <td> Stock Name: </td> <td> <?php $connect = mysql_connect("localhost","root", "") or die ("Couldn't Connect!"); mysql_select_db("stock", $connect) or die("Couldn't find db"); // select database $query=("SELECT id, stockname FROM stocks"); $result = mysql_query ($query); echo "<select name=stock value=''>Edit Stock QTY</option>"; while($nt=mysql_fetch_array($result)) { //Array or records stored in $nt echo "<option value=$nt[id]>$nt[stockname]</option>"; /* Option values are added by looping through the array */ } ?> </td> </tr> <tr> <td> Stock Qty: </td> <td> <input name="stock_qty1" type="text" /> </td> </tr> </table> <input name="submit" type="submit" value="Update stock items" /> </form> </td> </table> </td> </table> <BR /><BR /> <?php $connect = mysql_connect("localhost","root", "") or die ("Couldn't Connect!"); mysql_select_db("stock", $connect) or die("Couldn't find db"); // select database $query=("SELECT stockname, stockqty FROM stocks"); $result = mysql_query ($query); echo "<table border='1'>"; echo "<tr><th>Stock Name</th> <th>Stock Quantity</th></tr>"; while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<tr><td>"; echo $row['stockname']; echo "</td>"; echo "<td>"; echo $row['stockqty']; echo "</td>"; } ?> </center> <table> <tr> <td> </td> <td> </td> </tr> </body> </html> Here is the page that handles the form Code: [Select] <?php session_start(); $_SESSION['nt']; $sn = $_SESSION['nt']; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>stock</title> </head> <body> <?php $submit1 =&$_POST['submit1']; $stockname =&$_POST['stockname']; $stockqty =&$_POST['stockqty']; $submit = &$_POST['submit']; $stockqty1 = &$_POST['stock_qty1']; if(isset($submit1)) { if(isset($stockname) && ($stockqty)) { $connect = mysql_connect("localhost","root", "") or die ("Couldn't Connect!"); mysql_select_db("stock", $connect) or die("Couldn't find db"); // select database $query = mysql_query("SELECT * FROM stocks WHERE stockname='$stockname'"); $numrows = mysql_num_rows($query); if ($numrows!=0) { echo("Item already exists!");?><BR /><BR /><?php echo("Please Add A non-existent Item!");?> <a href="stockmanager.php">Try again!</a><?php } else { $queryreg = mysql_query("INSERT INTO `stocks` (stockname, stockqty) VALUES ('$stockname','$stockqty')"); echo("Stock has been added"); ?><BR /><BR /> <?php echo("Click here to return to stock manager!");?> <a href="stockmanager.php">Click Here!</a><?php } } else { echo("Please fill in all fields to add stock!"); } } if(isset($submit)) { $connect = mysql_connect("localhost","root", "") or die ("Couldn't Connect!"); mysql_select_db("stock", $connect) or die("Couldn't find db"); // select database $query=("SELECT id, stockname FROM stocks"); $result = mysql_query ($query); while($nt=mysql_fetch_array($result)) { //Array or records stored in $nt echo "<option value=$nt[id]>$nt[stockname]</option>"; /* Option values are added by looping through the array */ } $queryreg = mysql_query("SELECT * FROM stocks WHERE stockqty='$stockqty1'"); $numrows = mysql_num_rows($queryreg); $update = mysql_query("UPDATE stocks SET stockqty=$stockqty1 WHERE stockname=$sn'"); echo("Item has been updated"); } ?> </body> </html> Any help woul;d be really great, i ahve tried everything for days on end now Thanks Lance Ok, so I am trying to make a database with PHP and I hit a snag. First off, here is the code in question. Code: [Select] <?php //GET CONFIG include_once('../scripts/config.inc.php'); //SET VARIABLES TO SOMETHING A BIT SHORTER FOR THE DATABASE OPTIONS $host = $config['db_settings']['db_host']; $user = $config['db_settings']['db_user']; $pass = $config['db_settings']['db_pass']; $prefix = $config['db_settings']['db_prefix']; $db = $config['db_settings']['db_name']; $con = mysql_connect($host, $user, $pass); $select_db = mysql_select_db($db); $c_tables = ''; $c_database = "CREATE DATABASE IF NOT EXISTS $db;"; //NOW CONNECT TO THE SERVER if ($con){ echo('Connected Succesfully to the Server <br />'); } else die('Could not connect to database: ' . mysql_error()); //NOW CONNECT TO THE DATABASE if ($select_db){ echo('Selected Database Succesfully <br />'); } //IF DATABASE COULD NOT BE SELECTED THEN TRY TO MAKE IT else if(!$select_db){ echo('Could not select the databse. Trying to create it... <br />'); if(mysql_query($c_database)){ echo('Database has been created...<br />'); if($select_db){ echo('Database Has Been Selected.'); } else die('Can not select database!' . mysql_error()); } else die('Could not created the database!'); } else die ('Could not create or select the database!: ' . mysql_error() . '<br />'); //NOW CREATE THE TABLES mysql_close; ?>The problem is, that if I drop the database to test that it creates the database, it does indeed make it, but after it makes the database it will not select it until I reload the page and intern the script. So this is the output I would get. (The fact that there is no error is what is causing me to be confused) 1st load that makes new database: Quote Connected Succesfully to the Server Could not select the databse. Trying to create it... Database has been created... Can not select database! 2nd load that should not be needed to select the database: Quote Connected Succesfully to the Server Selected Database Succesfully |
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