CODEWITHSUNDEEP
phpjquerymysqljsonajax
cs-smith
cs-smith

Reputation: 25

ajax not responding with all array data

ajax comes back with 1 element of an array (data from mysql database) instead of all rows, pls help:

php code:

<?php
header("Content-type:application/json");
$srchString = trim($_POST['q']);
$sql = mysqli_query($connect,"SELECT * FROM dbname.tblname 
                       WHERE fieldname LIKE '%$srchString%' LIMIT 7");
confirmQuery($sql);
if(mysqli_num_rows($sql)>=1){
      $mem = mysqli_fetch_array($sql,MYSQLI_ASSOC);
      echo json_encode($mem);
}
?>

js:

$("#q").on("keyup",function(){
  var q = $(this).val();

  $.ajax({
    type: "POST",
    dataType:"json",
    url:"process.php",
    cache:false,
    data:"q="+ q,
    success:function(data){
      $.each( data, function( key, value ) {
        console.log(value.tel);
      });
    }
  });
});

Upvotes: 1

Views: 45

Answers (2)

madalinivascu
madalinivascu

Reputation: 32354

Use mysqli_fetch_all() 

if(mysqli_num_rows($sql)>=1){
      $mem = mysqli_fetch_all($sql,MYSQLI_ASSOC);
      echo json_encode($mem);
}

Upvotes: 1

mario.van.zadel
mario.van.zadel

Reputation: 2949

You have to iterate over all fetched rows and put them into an array. After that you are able to call json_encode to get a JSON array.

Please replace

if(mysqli_num_rows($sql)>=1){
      $mem = mysqli_fetch_array($sql,MYSQLI_ASSOC);
      echo json_encode($mem);
}

with

if(mysqli_num_rows($sql)>=1) {
    $rows = array();      
    while($row = mysqli_fetch_array($sql,MYSQLI_ASSOC)) {
        $rows[] = $row;
    }
    echo json_encode($rows);
}

Upvotes: 2

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