Question

我正在尝试使用Mysql和Ajax从表中返回数据，但变量中没有任何内容。连接到数据库是好的，查询也可以。

该表包含2个ID，我正在尝试检索它们：

process.php

<script>
function print() {

    $.ajax({  
         url: "process.php",   
         dataType: 'json',
         success : function(data)
         {
             console.log(data); // OUTPUT
         } 
    });
}
</script>

Ajax功能

Object {current_field: null, field_count: null, lengths: null, num_rows: null, type: null}

输出

 // When the user clicks anywhere outside of the modal, close it
window.onclick = function(event) {
 if (event.target == myModal) {
    myModal.style.display = "none";
}
}

Answer 1

PHP中的$ result变量实际上并没有返回ID，你必须遍历它才能获得如下ID：

<?php
    $con = mysqli_connect('localhost','root','root','DBB');
    if (!$con) {
        die('Could not connect: ' . mysqli_error($con));
    }


    $sql = "SELECT id FROM events";
    $ids = array();
    if ($result = mysqli_query($con,$sql)) {
        while ($obj = $result->fetch_object()) {
            $ids[] = $obj->id;
        }
    }

    echo json_encode($ids);
?>

Answer 2

你需要遍历每一行：

<?php
    $con = mysqli_connect('localhost','root','root','DBB');
    if (!$con) {
        die('Could not connect: ' . mysqli_error($con));
    }


    $sql = "SELECT id FROM events";
    $rows  = array();
    if ($result = mysqli_query($con,$sql)) {
        while ($row= $result->fetch_array(MYSQL_ASSOC)) {
            $rows[] = $row;
        }
    }

    echo json_encode($rows);

   $result->close();
   $con->close();
?>

返回数据Ajax - Mysql

2 个答案: