onclick event alert value of variable php mysql

Question

I want to alert id when user click on the button.But i am getting the same id on every button whereas when i inspect element it have different values. This is a part of my code

Fetching code:

$sql="select * from `table_name`";
$result=mysqli_query($con,$sql) or die(mysqli_error($con));
while($row=mysqli_fetch_array($result))
{
<button type="button" class="btn btn-danger" id="block" onclick="blockit(this.value);" value="<?php echo $row['id'];?>">Block</button>

Query:

<script>
function blockit()
{
var a=document.getElementById("block").value;
alert(a);
}
</script>

and your question is? also: you should note that the blockit-function you provided does not take any arguments... — Franz Gleichmann
– Franz Gleichmann, Commented Nov 23, 2016 at 10:19
@FranzGleichmann The question is i want to alert value that is related to that button. like if 1st result have value 1 thn it should alert 1 so on — A.R
– A.R, Commented Nov 23, 2016 at 10:21
what are you doing dude? if you are making function call like blockit(this.value) then use blockit(value) and alert(value)... simple — Happy Coding
– Happy Coding, Commented Nov 23, 2016 at 10:21

Happy Coding · Accepted Answer · 2016-11-23 11:01:05Z

2

pass the parameter to your custom function

function blockit(val)
{
  alert(val);
}

edited Nov 23, 2016 at 11:01

Happy Coding

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answered Nov 23, 2016 at 10:21

Sudharsan S

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Comments

Samir · Accepted Answer · 2016-11-23 10:28:32Z

0

For your every button you id is same. from line,

<button type="button" class="btn btn-danger" id="block" onclick="blockit(this.value);" value="<?php echo $row['id'];?>">Block</button>

Do remember, id should be unique. Since your id is same for all, it always consider only one, and it will neglect others. And hence you are getting same value.

Try with below code,

$sql="select * from `table_name`";
$result=mysqli_query($con,$sql) or die(mysqli_error($con));
while($row=mysqli_fetch_array($result))
{
<button type="button" class="btn btn-danger" onclick="blockit(this.value);" value="<?php echo $row['id'];?>">Block</button>

JS,

<script>
    function blockit(value){
       alert(value);
    }
</script>

edited Nov 23, 2016 at 10:28

answered Nov 23, 2016 at 10:22

Samir

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Comments

Dipen Desai · Accepted Answer · 2016-11-23 10:29:54Z

0

try this

$(document).ready(function(){
   $("#block").click(function(){
     var a=$(this).val();
     alert(a);
  });
});

answered Nov 23, 2016 at 10:29

Dipen Desai

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2 Comments

Jyothi Babu Araja Over a year ago

See how may buttons with same id as block

Dipen Desai Over a year ago

please change set class and use class in click event

Mr. Patel · Accepted Answer · 2016-11-23 10:36:37Z

0

Just change your BUTTON code as below,

while($row= mysqli_fetch_array($result))
{
      $ID= $row['id'];

  <button type="button" value="<?php echo $ID;?>" onclick="alert('<?php echo $ID;?>')" class="btn btn-danger" id="block" > Block</button>

You don't need to use blockit()

answered Nov 23, 2016 at 10:36

Mr. Patel

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Comments

Dipen Desai · Accepted Answer · 2016-11-23 10:38:27Z

0

<button type="button" class="btn btn-danger btn_block" onclick="blockit(this.value);" value="<?php echo $row['id'];?>">Block</button>

$(document).ready(function(){
 $(".btn_block").click(function(){
  var a=$(this).val();
  alert(a);
});
});

try this

answered Nov 23, 2016 at 10:38

Dipen Desai

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