ferellie
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Posts posted by ferellie
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[!--quoteo(post=351329:date=Mar 3 2006, 02:06 PM:name=craygo)--][div class=\'quotetop\']QUOTE(craygo @ Mar 3 2006, 02:06 PM) [snapback]351329[/snapback][/div][div class=\'quotemain\'][!--quotec--]
since web is a string you have to encase it in quotes
[code]$result = mysql_query("select * from $table WHERE categories LIKE '%web%' order by id desc limit $rows", $link)
or die ("Could not read data because ".mysql_error());[/code]
Ray
[/quote]
[!--quoteo(post=351329:date=Mar 3 2006, 02:06 PM:name=craygo)--][div class=\'quotetop\']QUOTE(craygo @ Mar 3 2006, 02:06 PM) [snapback]351329[/snapback][/div][div class=\'quotemain\'][!--quotec--]
since web is a string you have to encase it in quotes
[code]$result = mysql_query("select * from $table WHERE categories LIKE '%web%' order by id desc limit $rows", $link)
or die ("Could not read data because ".mysql_error());[/code]
Ray
[/quote]
thank you but how do I get the text from the input box to replace the text 'web' in my code?
Thanks -
I am using table with several fields which I display in a div no problem. I need to display all fields on a seperate page where the categories field contains the string web as an example. This is my code..
<?php
include "config.php";
// connect to the mysql server
$link = mysql_connect($server, $db_user, $db_pass)
or die ("Could not connect to mysql because ".mysql_error());
// select the database
mysql_select_db($database)
or die ("Could not select database because ".mysql_error());
// read data from database
$result = mysql_query("select * from $table WHERE categories LIKE %web% order by id desc limit $rows", $link)
or die ("Could not read data because ".mysql_error());
?>
AND TO DISPLAY IN A DIV
<?php
// print the data in a div
if (mysql_num_rows($result)) {
while ($qry = mysql_fetch_array($result)) {
?>
<div id="business">
<?php
print "$qry[business]<br>";
?>
</div>
<div id="businessdescription">
Business description:</div>
<?php
print "$qry[description]<br>";
?>
<div id="contactname">
Contact person:</div>
<?php
print "$qry[name]<br>";
?>
<div id="location">
Location:</div>
<?php
print "$qry[town]";
?>
<?php
print "$qry[postcode]<br>";
?>
<div id="telephone">
Telephone:</div>
<?php
print "$qry[phone]<br>";
?>
<div id="website">
<a href="
<?php
print "$qry[website]";
?>
">Visit Company Website</a></div>
<div id="endseperator"></div>
<?php
}
}
mysql_close();
?>
MY CONFIG.PHP CONTENTS IS
<?php
$server = "localhost"; // server to connect to.
$database = "pages"; // the name of the database.
$db_user = "pages"; // mysql username to access the database with.
$db_pass = "pass"; // mysql password to access the database with.
$table = "denb"; // database table
$rows = 500; // the number of table rows to display
?>
I would really appreciate any help,
Thank you.
using the LIKE command
in PHP Coding Help
Posted
foreach($_POST as $key => $val) { // This loops through the form variables
$$key = $val; // Assigns variables with the same names as the input names
} // i.e. the text box inputSearch would have a variable name of $inputSearch
// read data from database
$result = mysql_query("select * from $table WHERE keywords LIKE '%$inputSearch%' order by id desc limit $rows", $link)
or die ("Could not read data because ".mysql_error());
Problem is, the query results are not quite what I had in mind! If the field contained the words 'one two three' and the user searched for 'three two' the query would return nothing. Is there a way of modifying the search to accomplish this? I want my search to find any matching words.
Thanks again,
Peter