PHP - Graphically Displaying A Sqlite Database Through Php & Additional Languages
Hi everyone this is my first post so I'm sorry if I've posted it in the wrong section!
Before I begin I'd like to point out that I'm not just here to get someone to produce what I want for me! I have a genuine interest in this forum and I'm slightly surprised and disappointed in myself that I hadn't took the time to join sooner! Anyway to the point of this post: (Please bare with me I think I'll need to justify the use of SQLite) I am currently undergoing a University gorup project that uses wireless sensors to collect environmental data such as, Temperature, Humidity, Light levels and Dew point. This application is programmed in Python and it collects these pieces of data (every 30 seconds) and places them into an SQLite database. My job is to now graphically plot/show the data that the Python program collects and stores in the SQLite database on a web page. I'm using Ubuntu and have successfully installed LAMP, SQLite3 and the SQLite PD0 driver. I have also successfully established a connection to the SQLite database as well as displaying the data on a web page (locally - thats all I need!) in the form of a simple HTML table through the use of the following PHP script: Code: [Select] <?php try { //open the database $db = new PDO('sqlite:/var/databases/307Code/python/readings.db'); //now output the data to a simple html table... print "<table border=1>"; print "<tr><td>Date & Time Recieved</td><td>Node</td><td>Temp</td><td>Hum</td><td>Light</td><td>Dew</td></tr>"; $result = $db->query('SELECT * FROM readingstable'); foreach($result as $row) { print "<tr><td>".$row['Recieved']."</td>"; print "<td>".$row['Node']."</td>"; print "<td>".$row['Temp']."</td>"; print "<td>".$row['Hum']."</td>"; print "<td>".$row['Light']."</td>"; print "<td>".$row['Dew']."</td></tr>"; } print "</table>"; // close the database connection $db = NULL; } catch(PDOException $e) { print 'Exception : '.$e->getMessage(); } ?> What I'm now looking to do is the following: 1. Display the data from the SQLite database in the form of line graphs (I've seen that I may need to convert the data to XML?) 2. Make the application asynchronous, preferably every time a new entry is added to the SQLite database the line graph(s) update without a user needing to refresh the browser. 3. The SQLite database is storing a ridiculous floating point number for the time field in the database, for example: "2012-03-23 16:49:42.440818" is a entry in the SQLite database. Is there any way to omit the .440818 through the use of PHP? Or will I need to edit the Python script? Also one thing to note: The person in my group who built the Python script didn't build it to make the SQLite database give each database entry a unique ID/Primary key. Any tips/advice/help would be massively appreciated! Regards, Rich Similar TutorialsI am trying to convert and old website from mysql database to sqlite. One of the chores it must do is collect information from the database and put it in a list/menu select box on a page so the user can choose which item to pursue.
In the following (incomplete) snippit, I am doing something incorrectly because the sql query does get the proper information (I can put it in a table on the page just fine). But I'm having trouble getting the information into the select options on a list/menu. It appears to be putting them all, one after the other in the first option spot. The last one seems to be the only one of 6 or 7 that shows up.
It's been 10 or 12 years since I've messed with php, so I think I'm way behind... any help would be appreciated.
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<form action="sqlPropDisplay.php" method="post" id="Residential"> I have a commenting system and i have a limit of a certain number of comments to be shown. What i want to do is have a button on the bottom of the page at the end of the comments that are showing and when you click it ajax loads the next certain number of of rows (but not all of them),and then you click it again and it shows more of them, etc. So for example. comment 1 comment 2 comment 3 comment 4 --click button--(loads 4 more)--- comment 5 comment 6 comment 7 comment 8 --click button--(loads 4 more)-- comment 9 comment 10 comment 11 comment 12 etc. until there are no more rows. what's the best way to do this? (I know how to do the ajax and all, i just need help with the script to select the rows) Thanks. 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 wasn't sure where to ask this question, so I decided to ask here since it might require PHP alone to do the job. So, I recently changed from MySQL to SQLite ( Speed performance in the servers I run. (Not websites) ), so I was wondering if its possible to retrieve info from SQLite DB and display it on the website. I know how to do it with MySQL. The SQLite DB is located on an FTP server, so I will use ftp functions to get the data. So, if It's possible, can ya give me a small example on how to achieve this? Thanks. 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='" . $cols_row["colour_img_alt"] . "' /> "; } print " </td> </tr> "; } } else { echo "Aids!"; } print "</table>"; ?>] I am trying to display data from a database from a form entry here is the php <?php include('dbconnect.php'); @mysql_select_db($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 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.. 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 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> The following snippit should retrieve 10 images stored in an SQLite database table. But for some reason that mystifies me, it only retrieves 9. (Or one less than the number of images that are actually stored on the table):
___________
# retrieve photos for this property $query = "SELECT * FROM binary_data where propID = $prop_id"; $result = $db->query($query); $row = $result->fetch(PDO::FETCH_ASSOC); # save each image data to file foreach($result as $row) { $filename = $row['filename']; $image = $row['bin_data']; file_put_contents($filename, $image); } ?> _____________ It saves the ones it gets to the temporary files as it should, and they appear on the webpage as they should. But one is always missing. Any suggestions/corrections appreciated. -Robert Hi guys, i was wondering.. Languages? How does that excactly work? Where do you save the different language strings, and that stuff? Do you have php files' for each language, and just includes them? Or,.. Anyone who can explain me? thanks =) 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 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. 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 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/>'; } ?> 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 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 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 |