Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in C:\wamp\www\ofer

[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.

[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.

[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.

[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.

[vartol]

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

[karthik_ppts]

If this thread has been successfully resolved, please Mark this Thread as Solved