如何将 Mysql/php 数组格式化为 json？

发布于 2024-11-05 15:21:42 字数 382 浏览 0 评论 0原文

之前问过这个问题，但我已将问题范围缩小到这段代码。这是我的代码，当我运行它时，它只是说“null”..

$getmsg = "SELECT * FROM user WHERE account_id = $id";      
$showmsg = @mysqli_query ($dbc, $getmsg);
        while ($row = mysqli_fetch_array($showmsg, MYSQLI_ASSOC)) {

$arrResults = array($row['user_username']);


} // END WHILE


// Print them out, one per line
echo json_encode($arrResults);

原文

Asked this before, but I've narrowed down the issue to this bit of code. Here's my code, when I run it, it just says "null"..

$getmsg = "SELECT * FROM user WHERE account_id = $id";      
$showmsg = @mysqli_query ($dbc, $getmsg);
        while ($row = mysqli_fetch_array($showmsg, MYSQLI_ASSOC)) {

$arrResults = array($row['user_username']);


} // END WHILE


// Print them out, one per line
echo json_encode($arrResults);

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去了角落 2024-11-12 15:21:42

首先，您已将 echo 放在循环之外，该循环仅回显最后一项而不是所有人，并且您不会检查查询是否有错误。

相反，这就足够了：

$getmsg = "SELECT * FROM user WHERE account_id = $id";      
$result = @mysqli_query($dbc, $getmsg) or die("Error: " . mysql_error());
$result = mysql_fetch_assoc($result);
echo json_encode($result);

它将结果放入一个 assoc 数组中，然后将整个数组转换为 json 并打印它。

First of all you have put the echo outside the loop which just echoes the last item instead of everyone and you don't check if there is a error with your query.

Instead this would be sufficient:

$getmsg = "SELECT * FROM user WHERE account_id = $id";      
$result = @mysqli_query($dbc, $getmsg) or die("Error: " . mysql_error());
$result = mysql_fetch_assoc($result);
echo json_encode($result);

It puts the result in one assoc array and then converts the whole array to json and prints it.

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