Get data from PHP query using JSON not working

Question

my php code is:

$query = "SELECT * FROM users WHERE user='admin' AND password='MTIz'";
$result = $link->query($query);
$yes = array();
$yes[] = $result->num_rows;
echo json_encode($yes);

And my HTML code is:

  $.ajax({
    url: 'vlogin.php',
    type: 'POST',
    data: myData,
    dataType: 'json',
    contentType: "application/json; charset=utf-8",
    success: function(yes) {
    alert(yes.Result);}
});

Not return anything. What appens? Thanks

do a console.log(yes) to see exactly what's inside your response — Alex Tartan
– Alex Tartan, Commented Jan 21, 2016 at 10:40
What kind of Object is $link ? What is $result->num_rows supposed to contain? Is it an object? Is it an integer? Where do you define the Result index? When writing $yes = array(); $yes[] = $result->num_rows; you won't have any $yes['Result'] but a $yes[0]['Result'] — Dragos
– Dragos, Commented Jan 21, 2016 at 10:42

Lex Lustor · Accepted Answer · 2016-01-21 10:54:15Z

1

When using $.ajax(), I usually implement the error callback :

From the jquery doc : error (Type: Function( jqXHR jqXHR, String textStatus, String errorThrown ))

Implementing it allows you to see if the call is generating (bad JSON) or receiving (webservice error) errors : you can log/alert textStatus and errorThrown.

$.ajax({
      url: 'vlogin.php',
      type: 'POST',
      data: myData,
      dataType: 'json',
      contentType: "application/json; charset=utf-8",
      success: function(yes) {
          console.log(yes.Result);
      },
      error: function(jqXHR, textStatus, errorThrown) {
          console.log(textStatus);
          console.log(errorThrown);
      }
});

answered Jan 21, 2016 at 10:54

Lex Lustor

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Comments

nerdyDev · Accepted Answer · 2016-01-21 10:42:15Z

0

$query = "SELECT * FROM users WHERE user='admin' AND       password='MTIz'";
$result = $link->query($query);
$yes = array();
$yes[] = $result->num_rows;
echo json_encode($yes);
die;

apply die after the json_encode($yes); it will give you result then

and alert only yes like this alert(yes);

answered Jan 21, 2016 at 10:42

nerdyDev

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3 Comments

Dragos Over a year ago

Could you also please explain how does that die; help?

Jon Over a year ago

Because some times more code is ran before being sent to the next page.

nerdyDev Over a year ago

sometimes without die; the function after encoding direct to any other controller. i faced the same problem when i applied die;, the issue was resolved. not sure it will work for you guyz or not! but it worked for me.

Vipin Sharma · Accepted Answer · 2016-01-21 10:47:26Z

0

Try Below

 $query = "SELECT * FROM users WHERE user='admin' AND password='MTIz'";
 $result = $link->query($query);
 $yes = array();
 $yes['record'] = $result->num_rows;
 echo json_encode($yes);

And .ajax

  $.ajax({
          url: 'vlogin.php',
          type: 'POST',
          data: myData,
          dataType: 'json',
          contentType: "application/json; charset=utf-8",
          success: function(yes) {
          console.log(yes.record);
    }
 });

edited Jan 21, 2016 at 10:47

answered Jan 21, 2016 at 10:42

Vipin Sharma

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