Please help whats wrong?

Question

jakx12 0 Junior Poster

15 Years Ago

The code is this:

<html>
<head>
<link rel=stylesheet type=text/css href=style.css>
<title>
Search Results
</title>
</head>
<body>
<?php include("header.php"); ?>
<center>
<div class=indexboxlarge2>
<?php
SESSION_start();
include("dbcon.php");
$searchresult = $_POST['result'];
dbcon();
$query = mysql_query("SELECT * FROM user_info WHERE 1_name LIKE '%$searchresult%' OR WHERE 2_name LIKE '%$searchresult%'") or die(mysql_error());
$total = mysql_num_array($query);
if ($total!=0) {
while ($row = mysql_fetch_array($query)) {
echo $row["1_name"];
echo " ";
echo $row["2_name"];
echo "<br>";
echo "<br>";
}
}
else	
{
	echo "Sorry there results to display.";
}

?>
</div>
</center>
</body>
</html>

i get this error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE 2_name LIKE '%"the search result%'' at line 1

Were i wrote the search result is what the user input is.

Any help will be happily accepted,

jakx12

php

3 Contributors
7 Replies
137 Views
9 Hours Discussion Span
Latest Post 15 Years Ago Latest Post by uncle_smith

All 7 Replies

diafol

15 Years Ago

You may do well to escape your variable or "brace" it, e.g.

mysql_query("SELECT * FROM user_info WHERE 1_name LIKE '%{$searchresult}%' OR WHERE 2_name LIKE '%{$searchresult}%'")

OR

mysql_query("SELECT * FROM user_info WHERE 1_name LIKE '%" . $searchresult . "%' OR WHERE 2_name LIKE '%" . $searchresult . "%'")

There are also issues with the HTML - it seems to be poorly formed.

diafol

15 Years Ago

There's no such function as that. Do you want mysql_num_rows or mysql_fetch_array ?

I assume that this is not a new error, it's just that pHp stopped before this previously due to another error (your ' OR ' troubles).

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jakx12 0 Junior Poster · Answer 1 · 2009-01-27T02:57:27+00:00

thanks, but i get this now!

Fatal error: Call to undefined function mysql_num_array() in /home/a8896310/public_html/searchresult.php on line 18

jakx12 0 Junior Poster · Answer 2 · 2009-01-27T02:57:59+00:00

thanks but i get this now:

Fatal error: Call to undefined function mysql_num_array() in /home/a8896310/public_html/searchresult.php on line 18

jakx12 0 Junior Poster · Answer 3 · 2009-01-27T03:26:06+00:00

ok so i have tried most things here is the modded code, but i now get this error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/a8896310/public_html/searchresult.php on line 18

the code:

<html>
<head>
<link rel=stylesheet type=text/css href=style.css>
<title>
Search Results
</title>
</head>
<body>
<?php include("header.php"); ?>
<center>
<div class=indexboxlarge2>
<?php
SESSION_start();
include("dbcon.php");
$searchresult = $_POST['result'];
dbcon();
$query1 = mysql_query("SELECT * FROM 'user_info' WHERE '1_name' LIKE '%{$searchresult}%' OR WHERE '2_name' LIKE '%{$searchresult}%';");
$total = mysql_num_rows($query1);
if ($total != 0) {
while ($row = mysql_fetch_array($query1)) {
echo $row["1_name"];
echo " ";
echo $row["2_name"];
echo "<br>";
echo "<br>";
}
}
else	
{
	echo "Sorry there results to display.";
}

?>
</div>
</center>
</body>
</html>

sorry about this but its really annoying me!

diafol · Answer 4 · 2009-01-27T03:53:47+00:00

mysql_query("SELECT * FROM 'user_info' WHERE '1_name' LIKE '%{$searchresult}%' OR WHERE '2_name' LIKE '%{$searchresult}%';");

You've quoted your fieldnames - don't! Also you've done WHERE twice! And you've got double ';'!

mysql_query("SELECT * FROM user_info WHERE 1_name LIKE '%{$searchresult}%' OR 2_name LIKE '%{$searchresult}%'");

Try that.

uncle_smith 7 Light Poster · Answer 5 · 2009-01-27T04:24:46+00:00

The code is this:

<html>
<head>
<link rel=stylesheet type=text/css href=style.css>
<title>
Search Results
</title>
</head>
<body>
<?php include("header.php"); ?>
<center>
<div class=indexboxlarge2>
<?php
SESSION_start();
include("dbcon.php");
$searchresult = $_POST['result'];
dbcon();
$query = mysql_query("SELECT * FROM user_info WHERE 1_name LIKE '%$searchresult%' OR WHERE 2_name LIKE '%$searchresult%'") or die(mysql_error());
$total = mysql_num_array($query);
if ($total!=0) {
while ($row = mysql_fetch_array($query)) {
echo $row["1_name"];
echo " ";
echo $row["2_name"];
echo "<br>";
echo "<br>";
}
}
else    
{
    echo "Sorry there results to display.";
}

?>
</div>
</center>
</body>
</html>

i get this error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE 2_name LIKE '%"the search result%'' at line 1

Were i wrote the search result is what the user input is.

Any help will be happily accepted,

jakx12

You not supposed to have 2 'WHERE'
try this:

WHERE 1_name LIKE '%{$searchresult}%' OR 2_name LIKE '%{$searchresult}%'

and you may want to try to use backticks (key above the tab key) to quote the field names like this:
1_name

Please help whats wrong?

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