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PHP - Am Having Problem Displaying Image From My Database
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
Similar TutorialsHi, 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? Not really sure how to get the images I have stored in MySQL into a html form. I can call-up the text fields from the database but it cannot seem to find the index for the images. Here is my code:- <?php session_start(); mysql_connect("localhost","root","abc") or die ("Error! Cannot connect to database"); mysql_select_db("theimageworks") or die ("Cannot find database"); $query = "SELECT * FROM jobs"; $result = mysql_query($query) or die (mysql_error()); ?> <?php //display data in html table echo "<table>"; echo "<tr><td>Username</td><td align='center'>Message</td><td>Product Image</td></tr>"; while($row = mysql_fetch_array($result)) { echo "</td><td>"; echo $row['username']; echo "</td><td>"; echo $row['message']; echo "</td></tr>"; echo $row['image']; } echo "</table>"; ?> The error message I get is "Notice: Undefined index: image in....." Thanks in advance! 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 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 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.. 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>"; ?>] 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> 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 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> 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 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 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 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/>'; } ?> 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, This has been baffling me for a couple hours now and i cant seem to figure it out. I have some code which creates an array and gets info from a mysql database and then displays in a list. This works great but after adding more and more rows to my database the list is now becoming quite large and doesnt look great on my site. Is it possible to split the list into multiple columns of about 25 and if possible once 3 or 4 columns have been created start another column underneath. To help explain i would be looking at a layout as follows: Code: [Select] line 1 line 1 line 1 line 2 line 2 line 2 ... ... ... line 25 line 25 line 25 line 1 line 1 line 1 line 2 line 2 line 2 ... ... ... line 25 line 25 line 25Im guessing there should be some sort of if statement to check how many items are being displayed and to create a new column if necessary. Is this correct? Thanks, Alex Hey all, First off, if this is in the wrong section I apologize. I wasn't sure if it should be here or the mySQL section. What's going on is, I'm in the process of learning the Prepared Statement way of doing things and am changing / updating my code to reflect the changes. Everything was going fine until I attempted to do what I could do using old MySQL methods and that is display the queried results on the same page. I can place a query and display the results as they should be displayed if I only use one block of code. However, if I try to do any additional queries on the same page, they get killed and do not display anything even though I know the query is fine because I can test the exact same syntax below one a different page and it works. Here's a code snippet for an example: Code: [Select] Code: <table> <tr> <td> // The below code will display a selection box containing various strings such as "hello world", "great to be here", "Wowserz", "this is mind blowing" etc. that are stored in the database. <?php echo "<select = \"SpecialConditions\">"; if($stmt->num_rows == NULL){ echo "No results found."; }else{ while($stmt->fetch()){ echo "<option value=\"$specialId\">$specialcondition</option>"; } } echo "</select>"; ?> </td> <td> // If I place another fetch query below the above fetch() query, this one will not show up. This one is supposed to display values 1 - 20 that have been stored in the DB. <?php echo "<select = \"NumberSets\">"; if($stmt->num_rows == NULL){ echo "No results found."; }else{ while($stmt->fetch()){ echo "<option value=\"".$numbers."\">".$numbers."</option>"; } } echo "</select>"; ?> </td> </tr> </table> What am I doing wrong with this? When I use regular SQL queries I can display multiple results on the same page. The results are being pulled from two separate joined tables but I don't think that's the issue. Hello I'm trying to set up a user area for my site where it displays the current logged in users ranking and other information in the future.
<?
ini_set('display_errors', 1);
require_once "header.php";
$sql = "SELECT * FROM users WHERE username = ?";
if($stmt = mysqli_prepare($link, $sql)){
mysqli_stmt_bind_param($stmt, 's', $_SESSION['username']);
if(mysqli_stmt_execute($stmt)){
$info = mysqli_fetch_array($stmt);
echo "Current rank:" . $info['rank'];
} else {
echo "Can't find user";
}
}
mysqli_stmt_close($stmt);
?>
That's the code I currently have but it gives me the error "but get an error message of mysqli_fetch_array() expects parameter 1 to be mysqli_result" Hello freaks, Got a task here which is displaying correctly (I believe). I only have 3 data entries in the db right now. Like I said (I think) the display of the code below yields the right layout but it repeats the first db entry over and over. I got the display to work correctly like this: 1 2 3 4 5 6 7 8 9 I am trying to get this display result (so it is more eligable for the end user!) 1 4 7 2 5 8 3 6 9 Thanks in advance! CODE Code: [Select] <?php include_once "connect_to_mysql.php"; $cols = 3; $result = mysql_query("SELECT plantID, botanicalName FROM plants ORDER BY botanicalName"); $numrows = mysql_num_rows($result); $rows_per_col = ceil($numrows / $cols); $c = 1; $r = 1; while ($row = mysql_fetch_array($result)) { $plantID = $row["plantID"]; $botanicalName = $row["botanicalName"]; if ($r == $rows_per_col) { $c++; $r = 1; } else { $r++; } } $dyn_table = '<table width="750" cellpadding="0" cellspacing="0" border="0">'; for ($r = 1; $r <= $rows_per_col; $r++) { $dyn_table .= '<tr>'; for ($c = 1; $c <= $cols; $c++) { $dyn_table .= '<td><a href="plant_details.php?plantID = ' . $plantID . '" id="plantLink">' . $botanicalName . '</a></td>'; } $dyn_table .= '</tr>'; } $dyn_table .= '</table>'; ?> Hello - I'm opening my website up to visitors for free, and trying to bypass a login screen to go straight into the data content that was appearing after a user logged in. I have an index.php file that included the following code at the beginning: <?php session_start(); include("database.php"); include("login.php"); include("/vservers/skyranks/db_connect.php"); ?> <?header("Cache-control: private"); ?> <html> I deleted the "include("login.php"); line, and was successful at bypassing the username and login screen. However, the page that is supposed to display the data content is incomplete. In fact, it only displays my company's logo. Any ideas as to why the data content is not showing up? Thank you for any help with this, as my php is quite novice at this point. Regards - Joe |
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