How to pass string value to ajax in PHP & Mysql

Question

i need to show different values only in drop down list.drop down list selection based result cal in ajax and display the related data show below in the drop down list.

How to pass the string value to ajax in this script.

Anyone pls guide me

<!doctype html>
<html class="no-js" lang="en">
  <head>
    <script>
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>

    <div class="row">
      <div class="large-12 columns">
        <h1></h1>
      </div>
    </div>

    <div class="row">
      <div class="large-12 columns">
        <!--Main Tab Start-->
        <?php
        $res=mysql_query("select DISTINCT tag,id from password");

        if($res === false )
        {
            die(mysql_error());
        }
        ?>
        <form action="" method="post">
            <select name="tagg" onchange="showUser(this.value)">
            <option value="">Select Accounts</option>
        <?php
        while($row=mysql_fetch_array($res))
        {
            $dess=$row['tag'];
            $id=$row['id'];
            ?>
            <option value="<?php echo $id;?>"><?php echo $dess; ?></option>

            <?php
        }

        ?>
                    </select>
            </form>
             <div id="txtHint"><b>Person info will be listed here.</b></div>
        <!--Main Tab Ent-->
      </div>
    </div>
  </body>
</html>

Getuser.php

<?php
$q = intval($_GET['q']);
$result=mysql_query("select * from password WHERE id = '".$q."'");
if($result === false )
{
    die(mysql_error());
}
while($row=mysql_fetch_array($result))
{
    $desss=$row['tag'];
}
$res=mysql_query("select * from password WHERE tag = '".$desss."'");
if($res === false )
{
    die(mysql_error());
}
?>
<table>
<thead>
<tr>
<th>Id</th>
<th>Name</th>
<th>username</th>
<th>Password</th>
<th>Description</th>
<th>Link</th>
</tr>
</thead>
<?php

while($row=mysql_fetch_array($res))
{
    $id=$row['id'];
     $name=$row['name'];
    $url=$row['url'];
    $uname=$row['username'];
    $pass=$row['password'];
    $tag=$row['tag'];
    $des=$row['description'];

    ?>
    <tbody>
    <tr>
    <td><?php echo $id; ?></td>
    <td><?php echo $name; ?></td>
    <td><?php echo $uname; ?></td>
    <td><span data-tooltip aria-haspopup="true" class="has-tip" title="<?php echo $pass; ?>">View</span></td>
    <td><span data-tooltip aria-haspopup="true" class="has-tip" title="<?php echo $des; ?>">Description</span></td>
    <td><a href="<?php echo $url; ?>" title="<?php echo $name; ?>" target="_blank">Link</a></td>

    </tr>
    </tbody>
    <?php

}
?>
</table>

i need help how to pass the option value in string in this script . My script does not work to pass the string values .only work for int values. — Gobinath M
– Gobinath M, Commented Apr 17, 2015 at 6:55
Stop using MySQL it is deprecated, use instead MySQLi or PDO. Plain MySQL has security flaws and is no longer maintained - avoid it!! — Mm-Art-In
– Mm-Art-In, Commented Apr 17, 2015 at 7:13
@ Martin: Hi pls guide me you mention your security tips. i am a beginner. — Gobinath M
– Gobinath M, Commented Apr 17, 2015 at 7:16

krishna · Accepted Answer · 2015-04-17 07:24:55Z

4

I have run your code, it is working fine.There is only one problem ,if you want to pass string then change following code on your gestuser.php change

$q = intval($_GET['q']);

to

$q =($_GET['q']);

and change

<option value="<?php echo $id;?>"><?php echo $dess; ?></option>

to

<option value="<?php echo $dess;?>"><?php echo $dess; ?></option>

Other wise your code is working

edited Apr 17, 2015 at 7:24

krishna

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answered Apr 17, 2015 at 7:19

Sourabh

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2 Comments

halfer Over a year ago

@Gobinath: I hope you didn't take this advice - removing intval() will introduce a second SQL injection vulnerability into your code.

halfer Over a year ago

@Sourabh: can you fix this security problem in your answer? I suggest mysql_real_escape_string() here, even though - as the OP is aware - the library is deprecated.

Amrita Gupta · Accepted Answer · 2015-04-17 07:03:09Z

0

Edit your function as following and see what string you are passing through ajax

function showUser(str) {
    alert(str);
      if (str=="") {

edited Apr 17, 2015 at 7:03

answered Apr 17, 2015 at 6:49

Amrita Gupta

371 silver badge11 bronze badges

8 Comments

Gobinath M Over a year ago

Hi , i have copied getuser.php pls guide me

Nishant Solanki Over a year ago

@amrita this is not an answer.. this should be posted as a comment.. you can read further here for how to write an answer in stackoverflow stackoverflow.com/help/how-to-answer

Amrita Gupta Over a year ago

@NishantSolanki: I need 50 reputation to add a comment dats why

Nishant Solanki Over a year ago

@AmritaGupta oh ok.. but if you will post comments as answers than someone will downvote it and you will loose your reputations instead of gaining it :D, best luck ..

Gobinath M Over a year ago

@Amrita Gupta : Hi i try , i got an "int" value in alert box. i need help how to pass the string value to ajax

|

Nishant Solanki · Accepted Answer · 2015-04-17 07:10:45Z

0

the javascript ajax function fires when the select value has changed

<select name="tagg" onchange="showUser(this.value)">

now this.value means you are passing the selected value to the javascript function

<option value="<?php echo $id;?>"><?php echo $dess; ?></option>

you have to put string in your value attribute

something like

<option value="<?php echo 'YOUR STRING HERE';?>"><?php echo $dess; ?></option>

now in your getuser.php file your query is

"select * from password WHERE id = '".$q."'"

so you will have to change the query respectively

answered Apr 17, 2015 at 7:10

Nishant Solanki

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3 Comments

Gobinath M Over a year ago

Hi , i try to pass the string in option value but ajax cal not working only working for integer.

Nishant Solanki Over a year ago

as said by @Amrita what are you getting in alert box??

Gobinath M Over a year ago

Hi , i got an string value in alert box following your steps. but "getuser.php" i am try to echo the alert box value . now the value is "0" for all the drop down list selection.

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