help with error please

justAnoob · December 22, 2009

I posted this is the js section and I was told it was for the PHP section :confused:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING

<script language="JavaScript">
image0.src ="<?php echo $row['imgpath']; ?>"
</script

Deoctor · December 22, 2009

try using it like this

<script language="JavaScript">
image0.src ="<?php echo "$row['imgpath']"; ?>"
</script

trq · December 22, 2009

There is nothing syntactically wrong with that code, can we see more code surrounding?

trq · December 22, 2009

try using it like this

<script language="JavaScript">
image0.src ="<?php echo "$row['imgpath']"; ?>"
</script

Now, that IS syntactically incorrect.

Deoctor · December 22, 2009

try using it like this
<script language="JavaScript">
image0.src ="<?php echo "$row['imgpath']"; ?>"
</script
Now, that IS syntactically incorrect.

i know that it is wrong, as he doesnot gave more info..

trq · December 22, 2009

Why would you knowingly post incorrect code for someone to use? Makes no sense.

justAnoob · December 22, 2009

Just posting up the important sutff

I have one div which shows a full size image. And smaller divs which show the thumbnail. on mouseover of the thumbs, the full size is displayed in the bigger div . $row['imgpath'] should always show in the bigger div until mouseover of the other thumbs.

<?php
$sql = mysql_query("SELECT imgpath, imgpath2, imgpath3, imgpath4, imgpath5, thumb_1, thumb_2, thumb_3, thumb_4, thumb_5 FROM member_pictures WHERE thumb_1 = " . $_SESSION['picposted'] . "");
$result=mysql_query($sql);
?>

script language="JavaScript">
<!--
//thumbnail script
image0 =new Image();
image1 =new Image();
image2 =new Image();
image3 =new Image();
image4 =new Image();
// This defines the source of the display image 
image0.src ="<?php echo $row['imgpath']; ?>"
image1.src ="<?php echo $row['imgpath2']; ?>"
image2.src ="<?php echo $row['imgpath3']; ?>"
image3.src ="<?php echo $row['imgpath4']; ?>"
image4.src ="<?php echo $row['imgpath5']; ?>"
// This defines the source of the preview image 
document.images['pimage0'].src=image0.src;
document.images['pimage1'].src=image1.src;
document.images['pimage2'].src=image2.src;
document.images['pimage3'].src=image3.src;
document.images['pimage4'].src=image4.src;
// This defines what to do when an image is clicked on 
function image_click(clicks)
{
if(clicks==0){document.images['large'].src=image0.src;}
if(clicks==1){document.images['large'].src=image1.src;}
if(clicks==2){document.images['large'].src=image2.src;}
if(clicks==3){document.images['large'].src=image3.src;}
if(clicks==4){document.images['large'].src=image4.src;}
}
// --></script>

<div align="center"><img src="<?php echo "$row['imgpath']"; ?>" align="middle" border="0" width="380 height="225" name="large"></div>
<div align="center">
     <a onmouseover="javascript:image_click(0)"><img src="<?php echo "$row['thumb_1']"; ?>"  width="90" height="80" name="pimage0" border=0></a>
    </div>
   </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(1)"><img src="<?php echo "$row['thumb_2']"; ?>"  width="90" height="80" name="pimage1" border=0></a>
   </div>
.....................and so on

Deoctor · December 22, 2009

i think in this one

u have an error

<div align="center"><img src="<?php echo "$row['imgpath']"; ?>" align="middle" border="0" width="380 height="225" name="large"></div>
<div align="center">
     <a onmouseover="javascript:image_click(0)"><img src="<?php echo "$row['thumb_1']"; ?>"  width="90" height="80" name="pimage0" border=0></a>
    </div>
   </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(1)"><img src="<?php echo "$row['thumb_2']"; ?>"  width="90" height="80" name="pimage1" border=0></a>
   </div>
.....................and so on

remove the "" after echo..

like this

<div align="center"><img src="<?php echo $row['imgpath']; ?>" align="middle" border="0" width="380 height="225" name="large"></div>
<div align="center">
     <a onmouseover="javascript:image_click(0)"><img src="<?php echo $row['thumb_1']; ?>"  width="90" height="80" name="pimage0" border=0></a>
    </div>
   </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(1)"><img src="<?php echo $row['thumb_2']; ?>"  width="90" height="80" name="pimage1" border=0></a>
   </div>
.....................and so on

trq · December 22, 2009

All these....

<?php echo "$row['imgpath']"; ?>

should be....

<?php echo $row['imgpath']; ?>

Otherwise, I don't see anything wrong.

justAnoob · December 22, 2009

Here is the all the code. I just left out the CSS

<?php
error_reporting(E_ALL);
ini_set('display_errors',1);
session_start();
include_once 'header.php';
include "connection.php";
$sql = mysql_query("SELECT imgpath, imgpath2, imgpath3, imgpath4, imgpath5, thumb_1, thumb_2, thumb_3, thumb_4, thumb_5,FROM member_pictures WHERE thumb_1 = " . $_SESSION['picposted'] . "");
$result=mysql_query($sql);
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<script src="Scripts/AC_RunActiveContent.js" type="text/javascript"></script>

<script language="JavaScript">
<!--
//thumbnail script
image0 =new Image();
image1 =new Image();
image2 =new Image();
image3 =new Image();
image4 =new Image();
// This defines the source of the display image 
image0.src ="<?php echo $row['imgpath']; ?>"
image1.src ="<?php echo $row['imgpath2']; ?>"
image2.src ="<?php echo $row['imgpath3']; ?>"
image3.src ="<?php echo $row['imgpath4']; ?>"
image4.src ="<?php echo $row['imgpath5']; ?>"
// This defines the source of the preview image 
document.images['pimage0'].src=image0.src;
document.images['pimage1'].src=image1.src;
document.images['pimage2'].src=image2.src;
document.images['pimage3'].src=image3.src;
document.images['pimage4'].src=image4.src;
// This defines what to do when an image is clicked on 
function image_click(clicks)
{
if(clicks==0){document.images['large'].src=image0.src;}
if(clicks==1){document.images['large'].src=image1.src;}
if(clicks==2){document.images['large'].src=image2.src;}
if(clicks==3){document.images['large'].src=image3.src;}
if(clicks==4){document.images['large'].src=image4.src;}
}
// --></script>  
</head>

<body>
<table width="954" border="0" align="center">
<tr>
  <td height="702">
   <div id="PictureBox">
    <div id="divvy">
     <span class="bigger">
        Your picture has been posted -     </span>
     <span class="different">Picture Preview<span class="goto_around"> (goto </span></span><span class="style1"><span class="style3"><a href="http://www.mysite.com/example.php" class="goto">Active Pictures</a></span></span><span class="different"><span class="goto_around"> to add pictures)</span></span><br />
     <div id="postnew">
      <table width="705" bordercolor="#000000" bgcolor="#9AA4AD" id="table">
        <tr>
          <td>
          <table width="400" border="1" cellpadding="0" cellspacing="0" bgcolor="#FFFFFF">
  <tr>
    <td colspan="4"><div align="center"><img src="<?php echo "$row['imgpath']"; ?>" align="middle" border="0" width="380 height="225" name="large"></div></td>
  </tr>
  <tr> 
    <td colspan="4">
    <div align="center">
     Mouse over thumbnails to view larger image.
    </div>
    </td>
  </tr>
  <tr>
   <td bordercolor="#FFFFFF">
    <div align="center">
     <a onmouseover="javascript:image_click(0)"><img src="<?php echo "$row['thumb_1']"; ?>"  width="90" height="80" name="pimage0" border=0></a>
    </div>
   </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(1)"><img src="<?php echo "$row['thumb_2']"; ?>"  width="90" height="80" name="pimage1" border=0></a>
   </div>
  </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(2)"><img src="<?php echo "$row['thumb_3']"; ?>"  width="90" height="80" name="pimage2" border=0></a>
   </div>
  </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(3)"><img src="<?php echo "$row['thumb_4']"; ?>"  width="90" height="80" name="pimage3" border=0></a>
   </div>
  </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(4)"><img src="<?php echo "$row['thumb_5']"; ?>"  width="90" height="80" name="pimage4" border=0></a>
   </div>
  </td>
</tr>
</table> 
          
          
	  		  
           
          </td>
        </tr>
      </table>
     </div>
    </div>
   </div>  </td>
</tr>
</table>


</body>
</html>

Deoctor · December 22, 2009

change this

<td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(1)"><img src="<?php echo "$row['thumb_2']"; ?>"  width="90" height="80" name="pimage1" border=0></a>
   </div>
  </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(2)"><img src="<?php echo "$row['thumb_3']"; ?>"  width="90" height="80" name="pimage2" border=0></a>
   </div>
  </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(3)"><img src="<?php echo "$row['thumb_4']"; ?>"  width="90" height="80" name="pimage3" border=0></a>
   </div>
  </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(4)"><img src="<?php echo "$row['thumb_5']"; ?>"  width="90" height="80" name="pimage4" border=0></a>
   </div>
  </td>
</tr>
</table>

like

<td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(1)"><img src="<?php echo $row['thumb_2']; ?>"  width="90" height="80" name="pimage1" border=0></a>
   </div>
  </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(2)"><img src="<?php echo $row['thumb_3']; ?>"  width="90" height="80" name="pimage2" border=0></a>
   </div>
  </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(3)"><img src="<?php echo $row['thumb_4']; ?>"  width="90" height="80" name="pimage3" border=0></a>
   </div>
  </td>
  <td bordercolor="#FFFFFF">
   <div align="center">
    <a onmouseover="javascript:image_click(4)"><img src="<?php echo $row['thumb_5']; ?>"  width="90" height="80" name="pimage4" border=0></a>
   </div>
  </td>
</tr>
</table>

trq · December 22, 2009

If that is all the code, I don't see where $row is even defined. Is it?

justAnoob · December 22, 2009

just changed it whle you were probably typing

changed

$result=mysql_query($sql);

to this

$row = mysql_fetch_array($sql)

Now this.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

One day I think i understand it, next day I'm confused

trq · December 22, 2009

mysql_fetch_array() expects a result resource which is what mysql_query() returns.

if ($result = mysql_query($sql)) {
  if (mysql_num_rows($result)) {
    $row = mysql_fetch_array($result);
    // rest of your code should go here.
  }
}

justAnoob · December 22, 2009

wo, what? sorry.

trq · December 22, 2009

I'm not sure how it could be explained any simpler. mysql_fetch_array() expects a result resource which is what mysql_query() returns.

The ifs and check against mysql_num_rows() are there to prevent errors. You should always check your results before attempting to use them.

Deoctor · December 22, 2009

I'm not sure how it could be explained any simpler. mysql_fetch_array() expects a result resource which is what mysql_query() returns.

The ifs and check against mysql_num_rows() are there to prevent errors. You should always check your results before attempting to use them.

sorry i too didnt got it :confused:

so do mean that without if condition mysql_fetch_array() doesnot work

justAnoob · December 22, 2009

Actually, just the part where you say that the rest of the code goes here. I'm sorry, I'm not just looking for someone to give me the answer. I would still like to understand it.

trq · December 22, 2009

I'm not sure how it could be explained any simpler. mysql_fetch_array() expects a result resource which is what mysql_query() returns.

The ifs and check against mysql_num_rows() are there to prevent errors. You should always check your results before attempting to use them.

sorry i too didnt got it

so do mean that without if condition mysql_fetch_array() doesnot work

If there are no results there's no point passing an empty result resource to mysql_fetch_assoc().

trq · December 22, 2009

Actually, just the part where you say that the rest of the code goes here. I'm sorry, I'm not just looking for someone to give me the answer. I would still like to understand it.

Given my code above, the part where I sat '// rest of your code should go here'. That is the only place where $row is defined and guaranteed to have data.

justAnoob · December 22, 2009

got it. I think. I try a few things.

Deoctor · December 22, 2009

i think the error is in the query

$sql = mysql_query("SELECT imgpath, imgpath2, imgpath3, imgpath4, imgpath5, thumb_1, thumb_2, thumb_3, thumb_4, thumb_5 FROM member_pictures WHERE thumb_1 = " . $_SESSION['picposted'] . "");

try running the query in the mysql and see what result u get

by the way wht is the $_SESSION['picposted'] value

justAnoob · December 22, 2009

Nope, totally confused now. And wondering if the general way I have the code will even work at all, or if I should change everything.

Deoctor · December 22, 2009

Nope, totally confused now. And wondering if the general way I have the code will even work at all, or if I should change everything.

just try running the query in the mysql and see if u can get any results.

if not then use the code which has been given by thrope

trq · December 22, 2009

I don't see anything wrong with the query or the logic here. Whats the problem now?

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help with error please

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