Mr.Kiran Mr.Kiran - 16 days ago 5
Ajax Question

Handling ajax success response

PHP: lockState.php

require '../dbconn.php';
$query = mysql_query("select id, lockState,name from register_db where Id=1");
$items = array();
while ($row = mysql_fetch_object($query)) {
array_push($items, $row);
}
echo json_encode($items);


Result from query

[{"id":"1","lockState":"No","name":"Local Application"}]


Index.php

$.ajax({
type: "POST",
url: "feed/lockState.php",
data: ({id: 1}),
cache: false,
dataType:"json",
success: function (response) {
alert(JSON.stringify(response)); // [{"id":"1","lockState":"No","name":"Local Application"}]
alert(response.name); //***undefined***
if(response.name=='Local Application'){
callMyFunction(response.name);
}
},
error: function () {
alert("Oops..!! Something wrong!);

}
});


I'm totally lost where I'm doing wrong in using 'Success' response. Even I tried to
JSON.parse(response)
and tried to access the key:value, but still same
undefined
. Please help.

Answer

response[0].name will help you.

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