Question

我在这里做错了吗？请帮帮我......

example.html的

<script>
$.ajax({
      type: "POST",
      dataType: "json",
      data: {dat: a, pata: id},
      url: 'select.php',
      success: function(data) {
        alert(data.length); 
      }
    });
</script>

select.php

<?php
    $dbhost = "localhost";
    $dbuser = "root";
    $dbpass = "";
    $dbname = "empty_spot";


    $connection = mysqli_connect($dbhost,$dbuser,$dbpass,$dbname);

    $q1="select * from seats";

    $i = 0;
    $all_chairs = array();

    $query = mysqli_query($connection,$q1);

    while ($obj = mysqli_fetch_assoc($query))
    {
        $all_chairs[$i] = $obj["status"];
        $i++;
    }

    echo json_encode($all_chairs);

?>

Answer 1

你的问题就在这里把它data:{"dat":a,"pata":id}。如果a和id是变量，则定义它，否则也将它们放在引号

中

使用ajax从php文件中检索数组

1 个答案: