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. — CodeMan, Apr 17 '15 at 06:55
Stop using MySQL it is **deprecated**, use instead MySQLi or PDO. Plain MySQL has security flaws and is no longer maintained - avoid it!! — Martin, Apr 17 '15 at 07:13
@ Martin: Hi pls guide me you mention your security tips. i am a beginner. — CodeMan, Apr 17 '15 at 07:16
[Why shouldn't I use mysql_* functions in PHP?](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) — krishna, Apr 17 '15 at 07:20

score 4 · Accepted Answer · edited Apr 17 '15 at 07:24

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 '15 at 07:24

krishna

4,069
2
29
56

answered Apr 17 '15 at 07:19

Sourabh

500
2
14

@Gobinath: I hope you didn't take this advice - removing `intval()` will introduce a second SQL injection vulnerability into your code. – halfer Apr 20 '15 at 13:03
@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. – halfer Apr 20 '15 at 13:04

Amrita Gupta · Answer 2 · 2015-04-17T07:03:09.513

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 '15 at 07:03

answered Apr 17 '15 at 06:49

Amrita Gupta

37
1
11

Hi , i have copied getuser.php pls guide me – CodeMan Apr 17 '15 at 06:54
@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 http://stackoverflow.com/help/how-to-answer – Nishant Solanki Apr 17 '15 at 06:56
@NishantSolanki: I need 50 reputation to add a comment dats why – Amrita Gupta Apr 17 '15 at 07:00
@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 .. – Nishant Solanki Apr 17 '15 at 07:01
@Amrita Gupta : Hi i try , i got an "int" value in alert box. i need help how to pass the string value to ajax – CodeMan Apr 17 '15 at 07:09
@GobinathMahalingam: Is it the value of $id that you have passed ? – Amrita Gupta Apr 17 '15 at 07:12
@GobinathMahalingam: – Amrita Gupta Apr 17 '15 at 07:14
@GobinathMahalingam I strongly feel that you are usiing ajax too early.. you first need to clear corephp basic concepts.. – Nishant Solanki Apr 17 '15 at 07:17

score 0 · Answer 3 · answered Apr 17 '15 at 07:10

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 '15 at 07:10

Nishant Solanki

2,119
3
19
32

Hi , i try to pass the string in option value but ajax cal not working only working for integer. – CodeMan Apr 17 '15 at 07:13
as said by @Amrita what are you getting in alert box?? – Nishant Solanki Apr 17 '15 at 07:15
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. – CodeMan Apr 17 '15 at 07:22

How to pass string value to ajax in PHP & Mysql

3 Answers3