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PHP - display error if search unsuccessful

开发者 https://www.devze.com 2023-03-07 15:43 出处:网络
I\'m having trouble with my search script. Basically, all is fine with the search if the result is found but if there are no matches in the DB(MySQL) then my error doesn\'t display.. am i missing so

I'm having trouble with my search script.

Basically, all is fine with the search if the result is found but if there are no matches in the DB(MySQL) then my error doesn't display.. am i missing something? heres the code:

<?php
$term = $_GET['term'];
$sql = mysql_query("select * from search_e where content like '%$term%'");
while ($row = mysql_fetch_array($sql)){ 
$data = $row['content'];
$first_pos = strpos($data,$term);
if ($first_pos !== false) {
                  $output = substr($data,max(0,$first_pos - 100),200 + strlen($term));?>


<div>
<p class="ptitle"><?php echo $row["fn"]; ?></p><hr>
            Brief summary of contents:
            <hr class="hr">
            <p style="padding: 5px;">
        <i>"<?php echo $output; ?>" </i>..
            </p>


</div><br><br>
<?php
}
else  {?>
<div><?php echo "Sorry! No results were found using term: ".$_GET['term']."<br>Try using fewer Keywords"; ?></div>
<?php }?>
<?php
}
//close
    mysql_close();

?>

This may be something simple im doing wrong but i just cant figure it out. Also i know the code is dirty but its how i work.

I was also hoping to implement a little snippet i found browsing the net, which higlights specific words in a phrase.

function highlight($sString, $aWords) {
    if (!is_array ($aWords) || empty ($aWords) || !is_string ($sString)) {
        return false;
    }

    $sWords = implode ('|', $aWords);
    return preg_replace ('@\b('.开发者_Go百科$sWords.')\b@si', '<strong style="background-color:yellow">$1</strong>', $sString);
}

Is this possible to work into my script??


If I'm clear about what you're trying to accomplish, I would change it like so:

if(mysql_num_rows($sql) > 0) {
    while ($row = mysql_fetch_array($sql)) { 
        ...
    }
} else {
    echo("No Records!");
}

And barfoon is correct. Protect your web site and backend database from malicious users.

$term = mysql_real_escape_string($_GET['term']);

Edit

For completeness, after looking back over what you posted the reason you are getting no output is because if no matches are found anything inside of the while loop will not be executed, so your if($first_pos !== false) check is meaningless, except as a sort of 'sanity check' for records that did match.

To highlight the words using the function you posted, change:

<i>"<?php echo $output; ?>" </i>

To:

<i>"<?php echo highlight($output, array( $term )); ?>" </i>


Your logic is flawed:

  1. retrieve all rows in the database containing your search term
  2. loop over those rows:
    2a. retrieve a row
    2b. search the row AGAIN, using PHP, for the search term
    2c. display the content if it's found, or display an error if it's not. Do this for every row

Why have PHP re-search the content, when it's already been filtered by the database? As well, since the database query will not return any of the rows where you content does NOT appear, you'll never see the error message, as your php search will succeed each time.


the flow should basically be like this:

$term = mysql_real_escape_string($_GET['term']);

$sql = "SELECT .... WHERE content LIKE '%$term%'";
$result = mysql_query($sql) or die(mysql_error());

if (mysql_num_rows($result) == 0) {
   echo "Sorry, no results found"
} else {
   while ($row = mysql_fetch_assoc($result)) {
       ... do your output processing here ...
   }
}
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