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Editorial Team
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Asked: 2026-06-03T04:13:28+00:00

I am retrieving data from database using AJAX and PHP. In database one of

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I am retrieving data from database using AJAX and PHP. In database one of the columns contains path of an image folder. I am saving the value of path in a PHP variable named as $folder. This can be seen in getuser.php code.
I want this variable to be visible/available in one.php so that my images using this variable could be populated. How would i do this. I have tried including php as well but no use.

getuser.php

   <?php
$q=$_GET["q"];

$con = mysql_connect('localhost', 'san', '123');
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("holidayNet", $con);

$sql="SELECT * FROM image WHERE id = '".$q."'";

$result = mysql_query($sql);

echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Picture</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
    $folder = $row['FirstName'];
  echo "<tr>";
 echo "<td>" . $row['FirstName'] . "</td>";
  echo "<td>" . $row['LastName'] . "</td>";
  echo "<td>" . $row['Age'] . "</td>";
  echo "<td>" . $row['Hometown'] . "</td>";


  /*echo "<td>" . $row['Job'] . "</td>";*/
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);


?>

one.php

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>
<script type="text/javascript">
function showUser(str)
{
if (str=="")
  {
  document.getElementById("txtHint").innerHTML="";
  return;
  }
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>

<form>
<select name="users" onchange="showUser(this.value)">
<option value="">Select a person:</option>
<option value="1">Sn Qb</option>
<option value="2">Lois Griffin</option>
<option value="3">Glenn Quagmire</option>
<option value="4">Joseph Swanson</option>
</select>
</form>
<br />
<div id="txtHint"><b>Person info will be listed here.</b></div><br />

<img src="<?php echo $folder;?>/pic1.jpg" />
<img src="<?php echo $folder;?>/pic2.jpg" />
<img src="<?php echo $folder;?>/pic3.jpg" />
<img src="<?php echo $folder;?>/pic4.jpg" />


</body>
</html>
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1 Answer

  1. Editorial Team

    Hey you are creating a variable name $folder in the PHP file (getuser.php) which is getting called by AJAX. But its not available in the file name one.php.

    Only what you echo from getuser.php will be available in the JS variable xmlhttp.responseText

    So you will get all the Person Info echoed in the getuser.php, but wont get the $folder variable.

    Since you have specifically hardcoded 4 images. I suggest you to echo the img tags too in the getuser.php along with the other information from the database of the person selected.

    while($row = mysql_fetch_array($result))
{
    $folder = $row['FirstName'];
    echo "<tr>";
    echo "<td>" . $row['FirstName'] . "</td>";
    echo "<td>" . $row['LastName'] . "</td>";
    echo "<td>" . $row['Age'] . "</td>";
    echo "<td>" . $row['Hometown'] . "</td>";


    /*echo "<td>" . $row['Job'] . "</td>";*/
    echo "</tr>";
}
echo "</table>";

for($i = 0; $i < 4; $i++)
{
    echo '<img src="'.$folder.'/pic'.($i+1).'.jpg" />';
}

    And remove those image tags from the one.php page

    The other Solution:
    Suggestion which I can give is to add some separator to differentiate between the 2 things. One is the table which you want to print and the $folder variable value.
    For eg: consider separator ####

    Step 1:
    So now your code from the getuser.php will be 

    while($row = mysql_fetch_array($result))
{
    $folder = $row['FirstName'];
    echo "<tr>";
    echo "<td>" . $row['FirstName'] . "</td>";
    echo "<td>" . $row['LastName'] . "</td>";
    echo "<td>" . $row['Age'] . "</td>";
    echo "<td>" . $row['Hometown'] . "</td>";


    /*echo "<td>" . $row['Job'] . "</td>";*/
    echo "</tr>";
}
echo "</table>####".$folder;

    Step 2:
    Changes in the one.php to separate the 2 values

        if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
// split is the function which breaks the string with the value provided and then creates an array of the parts.
        var arrResponse = (xmlhttp.responseText).split("####");
        // added || below to validate the index 1 of arrResponse. if xmlhttp.responseText does not have #### it wont break the string and hence wont have the folderName part, in such cases javascript will give undefined value, by adding || we tell if value is present take the present value otherwise take it as an empty string ''
    var folderName = arrResponse[1] || ''
        document.getElementById("txtHint").innerHTML=arrResponse[0];

        // we will fetch all the image tags from your html page
        var arrImgs = document.getElementsByTagName('img');
        var imgCount = arrImgs.length;
        for(i = 0; i < imgCount; i++)
        {
            arrImgs[i].setAttribute('src', folderName+'/pic'+(i+1)+'.jpg');
        }
    }
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