sql json 不工作答案

【问题标题】：sql json not workingsql json 不工作
【发布时间】：2016-04-06 17:47:54
【问题描述】：

我创建了一个 php 脚本，它从数据库中获取值并将其保存在 json 中

$sql = "SELECT detail FROM BUNK WHERE id = '".$id."'";
$sql2 = "SELECT yes FROM BUNK WHERE id = '".$id."'";
$sql3 = "SELECT depend FROM BUNK WHERE id = '".$id."'";
$sql4 = "SELECT no FROM BUNK WHERE id = '".$id."'";

$result1 = mysqli_query($link,$sql);
$json['detail'] = $result1;
echo json_encode($json);

$result2 = mysqli_query($link,$sql2);
$json['yes'] = $result2;
echo json_encode($json);

$result3 = mysqli_query($link,$sql3);
$json['depend'] = $result3;
echo json_encode($json);

$result4 = mysqli_query($link,$sql4);
$json['no'] = $result4;
echo json_encode($json);

但我得到的结果是：

{"detail":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null}}{"detail":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null},"yes":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null}}{"detail":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null},"yes":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null},"depend":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null}}{"detail":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null},"yes":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null},"depend":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null},"no":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null}}

【问题讨论】：

php.net/manual/en/mysqli-result.fetch-assoc.php
为什么不在一个查询中完成这一切？
@chris85 new in phph.. :)
您只能执行一个查询（SELECT detail, yes, depend, no FROM ...），然后您必须获取该查询（请参阅these examples）。最后但同样重要的是，您只需编码一次，否则您的 JSON 对象无效。
非常感谢大家...！！

标签： php sql json

【解决方案1】：

基本：

$sql    = "SELECT detail, yes, depend, no FROM BUNK WHERE id = '".$id."'";
$result = mysqli_query( $link, $sql ) or die( $link->mysqli_error );
$rows   = mysqli_fetch_all( $result, MYSQLI_ASSOC );

$json   = json_encode( $rows );

这只是基本的，因为：

如果您的id 来自用户输入（即HTML 表单），您必须考虑prepared statements；
如果要自定义 JSON 字符串，则必须使用带有 mysqli_fetch_assoc 的循环，而不是使用 mysqli_fetch_all

$json 是生成的 JSON 字符串。输出它：

echo $json;

【讨论】：