jQuery/Php Error Illegal string offset from unserialized post data

user_7258982 0
Viewed 38

I'm posting a jquery serialize data posting from modal/pop up div tag with name attribute such as formdata[1]field_x

<div class="modal fade show" id="modal-payref-settings" data-bs-backdrop="static" data-bs-keyboard="false" tabindex="-1" aria-labelledby="modal-paycode-detailsLabel" aria-modal="true" role="dialog" style="padding-right: 17px; display: block;">
<div class="modal-dialog   modal-dialog-scrollableX   "><div class="modal-content ">
<div class="modal-body ">
<input type="text" name="formdata[1]payref_autonum" class="form-control dropdown-vw-item-text input-payref-autonum  ">
<input type="text" name="formdata[1]payref_prefix" class="form-control  input-payref input-payref-prefix  text-end">
<input type="text" name="formdata[1]payref_suffix" class="form-control  input-payref input-payref-suffix text-end">
<input type="text" name="formdata[1]payref_pad" class="form-control  input-payref input-payref-num-pad text-end" value="4">
<input type="text" name="formdata[1]payref_next" class="form-control  input-payref input-payref-num  text-end" value="876">
<input type="text" name="formdata[1]payref_sample" class="form-control input-payref-sample text-end" placeholder="Sample" aria-label="Sample" readonly="">
</div>
<!-- modal-body ends -->
</div>
</div>
</div>


$.ajax({
type: 'POST',
    url: 'model/trcode.php',
    data: {
        'a': 'SAVE_PAYREF',
    
      'formdata':$('#modal-payref-settings').find('[name^="formdata"]').serialize(),
    },
  //  dataType: "JSON",
    success: function (jsonStr) {
       alert(jsonStr);
    }
});

The Fetch/XHR capture the serialized posted data, which indicate the posting is correct. The same result return with print_r($_POST["formdata"] ); .

formdata%5B1%5Dpayref_autonum=Yes&formdata%5B1%5Dpayref_prefix=P2022&formdata%5B1%5Dpayref_suffix=&formdata%5B1%5Dpayref_pad=4&formdata%5B1%5Dpayref_next=876&formdata%5B1%5Dpayref_sample=

The data are getting un-serialized parse_str($_POST["formdata"], $formdata );. However $formdata return an empty array.

(
  [formdata] => Array
   (
     [1] => 
   )
)
php jquery json
2 Answers

You're trying to use a string as a full array

You can change the ajax code like

$.ajax({
type: 'POST',
    url: 'model/trcode.php',
    data: {
        'a': 'SAVE_PAYREF',
    
      'formdata':$('#modal-payref-settings').find('[name^="formdata"]').serialize(),
    },
  dataType: "JSON",
contentType: false,
            cache: false,
            processData:false,
    success: function (jsonStr) {
       console.log(jsonStr);
    }
});

And inside model/trcode.php page you can pass the

echo json_encode($_POST["formdata"]);

if use alert(jsonStr) instead of console.log(jsonStr) you will not get actual output sometimes because you are passing the object.

user_191266690

The issue is your form fields name attribute is not an array,(which you are thinking is an array, but has a syntax error).To fix this do like below:

change names like this: formdata[1][payref_autonum] and so on for others.

user_42483280
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