Json 响应只有值，但需要键 + 值答案

【问题标题】：Json response only value but key + value neededJson 响应只有值，但需要键 + 值
【发布时间】：2014-12-03 19:32:10
【问题描述】：

以下 PHP 代码仅翻译值，但我还需要每个值的键：

应该如何（期望的输出）：

[{"ID":"1","styleNo":"1","styleName":"Casual","placeholder":"Left Initial","type":"text","maxlength":"1","size":"small","position":"1"},{"ID":"2","styleNo":"1","styleName":"Casual","placeholder":"Right Initial","type":"text","maxlength":"1","size":"small","position":"2"},{"ID":"3","styleNo":"1","styleName":"Casual","placeholder":"Center Initial","type":"text","maxlength":"1","size":"small","position":"3"}]

实际情况如何（我不知道不同的括号是否重要？）：

[["1","1","Casual","Left Initial","text","1","small","1"],["2","1","Casual","Right Initial","text","1","small","2"],["3","1","Casual","Center Initial","text","1","small","3"]]

<?php
$style = $_GET['style'];
define('HOST', 'localhost');
define('USER', 'root');
define('PASS', '');
define('DBNAME', 'inscribe');

$db = new mysqli(HOST, USER, PASS, DBNAME);

if (mysqli_connect_errno()) {
    printf("Connect failed: %s<br/>", mysqli_connect_error());
}

$sql = "SELECT * FROM inputs WHERE styleNo = '".$style."'";
$result_db = $db->query($sql);
$all_result = $result_db->fetch_all();
echo json_encode($all_result);
$db->close();
?>

【问题讨论】：

那是什么问题？ $all_result 编码后是什么样子的？
它看起来像上面的“实际情况”部分，只打印值而不打印 ID、styleNo、styleName 等键...

标签： php mysql arrays json

【解决方案1】：

尝试在fetch_all() 中将mysql fetch 标志显式设置为MYSQLI_ASSOC，因为它的默认值为MYSQLI_NUM，它返回数字索引而不是关联索引。

例子：

$all_result = $result_db->fetch_all(MYSQLI_ASSOC);
echo json_encode($all_result);

旁注：转义输入或使用准备好的语句：

$style = $db->real_escape_string($_GET['style']);
$sql = "SELECT * FROM inputs WHERE styleNo = '$style' ";
$result_db = $db->query($sql);
$all_result = $result_db->fetch_all(MYSQL_ASSOC);
echo json_encode($all_result);

或者

$sql = "SELECT * FROM inputs WHERE styleNo = ?";
$select = $db->prepare($sql);
$select->bind_param('s', $_GET['input']);
$select->execute();
$result = $select->get_result();
$all_result = $result->fetch_all(MYSQL_ASSOC);
echo json_encode($all_result);

【讨论】：

@Marius 我很高兴这有帮助

【解决方案2】：

它正在做它应该做的事情。 sql查询返回结果行，这就是它给你的。快速解决方法是

$tmp = array();
$keys = array("ID", "styleNo", "placeholder", "type", "maxlength", "size", "position");
foreach($all_result as $row){
     $tmp[] = array_combine($keys, $row);
}

echo json_encode($tmp);

如果您的选择发生变化，这可能会中断。

【讨论】：