php - 如何通过ajax将mysql结果作为jSON传递

Question

我不确定如何通过 ajax JSON 将 mysql 查询的结果传递到 html 页面。ajax2.php

$statement = $pdo - > prepare("SELECT * FROM posts WHERE subid IN (:key2) AND Poscode=:postcode2");
$statement - > execute(array(':key2' => $key2, ':postcode2' => $postcode));
// $row = $statement->fetchAll(PDO::FETCH_ASSOC);
while ($row = $statement - > fetch()) {
    echo $row['Name']; //How to show this in the html page?
    echo $row['PostUUID']; //How to show this in the html page?
    $row2[] = $row;
}
echo json_encode($row2);

如何将上面的查询结果通过下面的ajax传递到html页面中显示？

我的 Ajax

$("form").on("submit", function () {
    var data = {
        "action": "test"
    };

    data = $(this).serialize() + "&" + $.param(data);
    $.ajax({
        type: "POST",
        dataType: "json",
        url: "ajax2.php", //Relative or absolute path to response.php file
        data: data,
        success: function (data) {
            //how to retrieve the php mysql result here?
            console.log(data); // this shows nothing in console,I wonder why?
        }
    });
    return false;

});

Answer 1

你的 json 编码应该是这样的:

 $json = array();
 while( $row = $statement->fetch()) {
     array_push($json, array($row['Name'], $row['PostUUID']));
 }

    header('Content-Type: application/json');
    echo json_encode($json);

在你的 javascript 部分，你不需要做任何事情来取回你的数据，它存储在 success 函数的数据变量中。您可以直接显示它，并用它在您的网页上做任何您想做的事情