Member Avatar for vartol

The code:

<link href="css/main_stylesnew.css" rel="stylesheet" type="text/css" />
<TR>
        <TD COLSPAN=2 background ="images/207_13.gif" WIDTH=53 HEIGHT=335 ALT="">
            </TD>
  <TD COLSPAN=5 align="center" valign="top" class="smallheader">



        <div id="google_translate_element"></div><script>
function googleTranslateElementInit() {
  new google.translate.TranslateElement({
    pageLanguage: 'ro',
    includedLanguages: 'en,it,es',
    layout: google.translate.TranslateElement.InlineLayout.SIMPLE
  }, 'google_translate_element');
}
</script><script src="//translate.google.com/translate_a/element.js?cb=googleTranslateElementInit"></script>
</br>
</br>
       Afisare Oferte Munca
        <hr />
        <br />
        <?php
        error_reporting(0);
        $search=$_REQUEST['search'];
        ?>
        <table width="414" border="0" cellpadding="5" cellspacing="0">
          <tr>
            <td width="19"><strong>Nr.</strong></td>
            <td width="77"><strong>Specialitate</strong></td>
            <td width="64"><strong>Compania</strong></td>
            <td width="64"><strong>Localitate</strong></td>
            <td width="128"><strong>Data introducerii</strong></td>
          </tr>
          <?php
          $sql="select * from oferte_munca where specialitate like '%".$_GET['search']."%' or localitate like '%".$_GET['search']."%' or iuser like '%".$_GET['search']."%' or tara like '%".$_GET['search']"%'order by date1";
          $result=mysql_query($sql);
          $i=0;
          while($row=mysql_fetch_array($result))
          {$i++;
          if($i%2==0)
          {$bgcolor="#FFFFFF";}
          else
          {$bgcolor='#C5CBC7';}
          print"<tr>";
            print"<td bgcolor=".$bgcolor."><a href=oferte.php?num=".$i."&search=".$search.">".$i."</a></td>";
            print"<td bgcolor=".$bgcolor."><a href=oferte.php?num=".$i."&search=".$search.">".$row['specialitate']."</a></td>";
            print"<td bgcolor=".$bgcolor."><a href=oferte.php?num=".$i.">".$row['iuser']."</a></td>";
            print"<td bgcolor=".$bgcolor.">".$row['localitate']."</td>";
             print"<td bgcolor=".$bgcolor.">".$row['tara']."</td>";
            print"<td bgcolor=".$bgcolor."><a href=oferte.php?num=".$i.">".$row['date1']."</a></td>";
          print"</tr>";
          }
          ?>
    </table>  </TD>

    </TR>

Thanks.

Edited by TrustyTony because: fixed formatting
  • 2 Contributors
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  • 8 Hours Discussion Span
  • Latest Post Latest Post by karthik_ppts
Member Avatar for karthik_ppts

Post your code with code tags and mention the line no that you get the error.
This will help us to identify the problem easily.

Member Avatar for vartol

The line 36 is this one that is indicated as the parse error line:
<?php $sql="select * from oferte_munca where specialitate like '%".$_GET."%' or localitate like '%".$_GET."%' or iuser like '%".$_GET."%' or tara like '%".$_GET"%'order by date1";
$result=mysql_query($sql);... ?>
Thanks.

Edited by vartol because: n/a
Member Avatar for karthik_ppts

Replace your line 36 with this

$sql="select * from oferte_munca where specialitate like '%".$_GET['search']."%' or localitate like '%".$_GET['search']."%' or iuser like '%".$_GET['search']."%' or tara like '%".$_GET['search']."%' order by date1";

Yo have missed '.' in tara like '%".$_GET."%' order by date1";

please use code tags to post your code and mark as solved if your problem solved.

Edited by karthik_ppts because: n/a
Member Avatar for vartol

Ok it solved the problem thank u so much for your support.

Member Avatar for karthik_ppts

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