"error in your SQL syntax"

[mikebyrne]

Im getting the error and cant spot my mistake

 

error in the queryYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'LIKE '%%'' at line 1

 

<?PHP
include("adminconnect.php");
$tbl_name = "product";

$cat = mysql_real_escape_string($_POST['cat']);
$input =
mysql_real_escape_string($_POST['searchfield']);

$query = "SELECT * FROM $tbl_name WHERE $cat LIKE '%$input%'";
$result = mysql_query($query) or die ("error in the query" . mysql_error());
if (mysql_num_rows($result) > 0) {
while($row = mysql_fetch_array) {
         $product = $row['product'];
         echo "$product was found<br>";
     }
} else {
    echo "No search results found";
}
?>
<select name="cat" onChange="setAction(this.options[this.selectedIndex].value);"> 
<option value="cd" selected="selected" >CD</option>
<option value="dvd" >DVD</option>
<option value="game" >Game</option>
</select>
<form name="ex2" method="get" action="searchresults.php">

<input type="text" name="searchfield" size="22" maxlength="40" id="srchdrop" />
<input type="submit" value="Search Now! »" id="gosrch" />

</td></tr>
</form>

[mkoga]

try echoing out the value of $query to get a better idea of the problem.

[mikebyrne]

Would that be echo "['$query']";?

[mikebyrne]

Ive tried this but it doesnt seem to be showing the varible but the page wont load so the problem seems to be with the line

 

$query = "SELECT * FROM $tbl_name WHERE $cat LIKE '%$input%'";

 

 

 

<?PHP
include("adminconnect.php");
$tbl_name = "product";

$cat = mysql_real_escape_string($_POST['cat']);
$input =
mysql_real_escape_string($_POST['searchfield']);

$query = "SELECT * FROM $tbl_name WHERE $cat LIKE '%$input%'";
$result = mysql_query($query) or die ("error in the query" . mysql_error());
echo $query;
if (mysql_num_rows($result) > 0) {
while($row = mysql_fetch_array) {
         $product = $row['product'];
         echo "$product was found<br>";
     }
} else {
    echo "No search results found";
}
?>
<select name="cat" onChange="setAction(this.options[this.selectedIndex].value);"> 
<option value="cd" selected="selected" >CDs</option>
<option value="dvd" >DVDs</option>
<option value="game" >Games</option>
</select>
<form name="ex2" method="get" action="searchresults.php">

<input type="text" name="searchfield" size="22" maxlength="40" id="srchdrop" />
<input type="submit" value="Search Now! »" id="gosrch" />

</td></tr>
</form>
</table>

 

[moon 111]

Check if its empty

 

<?php
if(empty($query))
  echo "Empty!";
?>

[mikebyrne]

Well it would be empty because the page wont load it just give me the error messege

[Catfish]

Try changing

 

$query = "SELECT * FROM $tbl_name WHERE $cat LIKE '%$input%'";

 

to

 

$query = "SELECT * FROM $tbl_name WHERE $cat LIKE '%".$input."%'";

 

if that doesn't work I would be monitoring the value of $input through your script to see what value is passed into the $query string.

[thenature4u]

while($row = mysql_fetch_array) {

 

change this to

 

while($row = mysql_fetch_array($result)) {

[mkoga]

Would that be echo "['$query']";?

 

no just echo $query.

[mikebyrne]

No, still getting the same error and tried Everything!!!!

 

I can pm people my complete code if you'd like to take a look?

[mikebyrne]

It has to be something to do with the placment of the php rather than the actual code?

[AndyB]

<form name="ex2" method="get" action="searchresults.php">

 

$cat = mysql_real_escape_string($_POST['cat']);

 

$input = mysql_real_escape_string($_POST['searchfield']);

 

Since the form method is GET there aren't going to be any POST variables.

 

Your form is also mangled. The select option needs to be inside the form tags. The onChange serves no purpose then.