[RESOLVED] SQL query is giving me an Parse error

[Danh]

Hi,

I am new to php and I ran into a problem. I got a tutorial on using php to search MYSQL database and this is the code they gave me.

$query = "select * from the_table where 1st_field like \"%$trimmed%\" order by 1st_field";

//EDIT HERE and specify your table and field names for the SQL query

I changed the_table to part_number (which is the name the table) and changed 1st_field to Part_Number (which is the field name) so the code looked like this

$query = "select * from part_number where Part_Number like \"%$trimmed%\" order by Part_Number";

I got this error: Parse error: parse error, unexpected T_STRING, expecting ',' or ';' in /tmp/disk/home/tutorial/Files/WWW/search.php on line 30

Can anybody help me with this? Thanks.

[Roger_Ramjet]

The problem is with your string quoting syntax - if that is line 30.

[Danh]

Thanks Roger, That fixed it.
I have one more problem with it thats left me scratching my head. Its giving me a parse error on line 105. Its the last line in the code which is ?> . Can anybody tell me what I am doing wrong.

My last 6 lines:

$a = $s + ($limit);
if ($a > $numrows) { $a = $numrows; }
$b = $s + 1;
echo "<p>Showing results $b to $a of $numrows</p>

?>

Gives me parse error:

Parse error: parse error, unexpected $ in /tmp/disk/home/tutorial/Files/WWW/search.php on line 105

Thanks,
Daniel

[laserlight]

Your have a string that is not properly delimited, it should be:

echo "<p>Showing results $b to $a of $numrows</p>";

[Danh]

Sorry, I forgot to add the quotation mark when I was typing. I do have it in the code that way.

[laserlight]

What is your actul code? Do not retype your code, copy and paste instead. If not, people will help you fix bugs that did not even exist.

[bradgrafelman]

When posting PHP code, please use the board's [PHP][/PHP] bbcode tags - they make the code much, much easier to read (as well as aiding the diagnosis of syntax errors). You should edit your post now and add these bbcode tags to help others in reading/analyzing your code.

[Danh]

Alright, here's the code that I am having the error with. Any help would be greatly appreciated.

<?php 

 // Get the search variable from URL

 $var = @$_GET['q'];
 $trimmed = trim($var); //trim whitespace from the stored variable

// rows to return
$limit=10;

// check for an empty string and display a message.
if ($trimmed == "")
 {
 echo "<p>Please enter a search...</p>";
 exit;
}	 

//check for a search parameter
if (isset($var))
 {
 echo "<p>We do not seem to have a search parameter!</p>";
 exit;
}

//connect to your database
include 'library/config.php';
include 'library/opendb.php';

//Build SQL Query
$query = 'select * from part_number where Part_Number like \"%$trimmed%\ order by Part_Number'; 

$numresults=mysql_query($query);
$numrows=mysql_num_rows($numresults);

//if we have no results offer addpartnumber

if ($numrows == 0)
 {
 echo '<h4>Results</h4>';
 echo '<p>Sorry, your search: &quot;" . $trimmed . "&quot; returned zero results</p>';

// addpartnumber
 echo '<p><a href=\"addpartnumber.php="	. $trimmed . "\" target=\"_blank\" title=\"Look up " . $trimmed . " on Google\">Click here</a>to try the search on google</p>'; 
}

// next determine if s has been passed to script, if not use0
 if (empty($s)) {
 $s=0;
}

//get results
 $query .= ' limit $s,$limit';
 $result = mysql_query($query) or die('Could not execute query');

//display what the person searched for
echo '<p>You searched for: &quot;" . $var . " &quot;</p>';

// begin to show results set 
echo 'Results';
$count = 1 + $s ;

//now you can display the results returned
 while ($row= mysql_fetch_array($result)) {
 $title = $row["Part_Number"];

 echo '$count.)&nbsp;$title' ;
 $count++ ;
}

$currPage = (($s/$limit) + 1);

//break before paging
 echo '<br />';

 // next we need to do the links to other results
 if ($s>=1) { // bypass PREV link if s is 0
 $prevs=($s-$limit);
 print '&nbsp;<a href=\"PHP_SELF?s=$prevs&q=$var\">&lt;&lt;
 $prevs 10</a>&nbsp&nbsp;';
}
// calculate number of pages needing links
 $pages=intval($numrows/$limit);

// $pages now contains int of pages needed unless there is a remainder from division

 if ($numrows%$limit) {
 // has remainder so add one page
 $pages++;
}

// check to see if last page
 if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {

 //not last page so give NEXT link
 $news=$s+$limit;

 echo "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</a>";
 }

$a = $s + ($limit);
 if ($a > $numrows) {$a = $numrows;}
 $b = $s + 1;
 echo "<p>Showing results $b to $a of $numrows</p>";

?>

Parse error: parse error, unexpected $ in /tmp/disk/home/tutorial/Files/WWW/search.php on line 105

[Roger_Ramjet]

Look at the colours!!

[Danh]

I am using Crimson Editor. It points things out but not the way it did above. Is there another program that makes things alot easier?

I corrected the code as I updated the one above and now I am getting " Please enter a search... " now I know that its suppose to say that when there's a empty string but I know what I am entering into the search is in my database (MySQL). What could I be doing wrong now?

[Roger_Ramjet]

Loads of things. Most particularly: suppressing errors while developing and debugging.

[bradgrafelman]

One SQL syntax error that jumps out at me is found here:

$query = 'select * from part_number where Part_Number like \"%$trimmed%\ order by Part_Number'; 

1. Why are you trying to escape a double quote here? You used single quotes as your string delimiter...
2. There's no ending quote for the LIKE string.
3. Since you used single quotes as your string delimiter, the $trimmed variable reference won't be parsed - it'll be left as the literal string '$trimmed'.
4. You shouldn't ever use 'SELECT *' - name the columns (and only the columns) you're actually going to pull data from.

[shu]

Dude, looks like you need a serious tutorial on proper use of single quotes (') and double quotes ("). I am surprise that your code actually give you any meaningful results at all, judging from the plethora of statements that the intended values of variables are not parsed by the PHP engine.

1) Use single quotes (') when dealing with literal strings only, eg:
echo '<p>Please enter a search...</p>';
Statement like $query .= ' limit $s,$limit'; will not give you the result desired, used double quotes (") instead.

2) Use double quotes when variables are involved, eg:
$query = "select * from part_number where Part_Number like '%$trimmed%' order by Part_Number";

Majority of your statements are bad!!!