如何将数据返回到jquery ajax回调函数？

Question

如何将数据返回到jquery ajax回调函数，其中数据来自PHP mysql查询..

<script language="javascript" type="text/javascript">
   $(document).ready(function() {
        $.post("getMySqlData", function(data){

        });
   });
</script>

这是执行mysql查询以从mysql数据库中选择数据的PHP代码

  <?php
    $host = "127.0.0.1";
    $user = "root";
    $password = "problem2";
    $database = "scsi_test_log";
    $cxn = mysqli_connect($host, $user, $password, $database) or die ("couldnt connect to server");

    $query = "SELECT test_header FROM scsi_test";
    $result = mysqli_query($cxn, $query) or die ("Couldn't execute query.");
  ?>

Answer 1

将结果提取到数组中，并在JSON encoding;

中回显

$rows = array();
while ( $row = mysqli_fetch_assoc( $result ) ) {
    $rows[] = $row;
}

echo json_encode( $rows );