如何编码多行的json? [英] How to encode json with multiple rows?
问题描述
在开始之前,我已经浏览了多个平台上的其他示例和Q& A,但它们似乎都无法解决我的问题.我试图通过json从MySQL返回多行.但是,我一直做不到.下面的代码显示了我的尝试.
Before I begin, I have looked through other examples and Q&A's on multiple platforms but none of them seem to solve my problem. I am trying to return multiple rows from MySQL via a json. However, I have been unable to. The code below shows my attempt.
我通过邮递员得到我的回复.第一个while
仅返回数据库中的最后一个条目,而do-while
返回所有条目,但未正确编码json,因为json输出syntax error
,但html部分显示了所有条目.
I get my responses via Postman. The first while
returns only the last entry in the database, and the do-while
returns all entries but doesn't encode the json properly, as the json outputs syntax error
but the html part shows all entries.
<?php
$dashboard_content_token = $_REQUEST["dashboard_content_token"];
$token = "g4";
require(cc_scripts/connect.php);
$sql = "SELECT * FROM `dashboard_content`";
$check = strcmp("$token", "$dashboard_content_token");
$statement = mysqli_query($con, $sql);
if (check) {
$rows = mysqli_fetch_assoc($statement);
if (!$rows) {
echo "No results!";
} else {
while ($rows = mysqli_fetch_assoc($statement)) {
$news_id = $rows['news_id'];
$image_url = $rows['image_url'];
$news_title = $rows['news_title'];
$news_description = $rows['news_description'];
$news_article = $rows['news_article'];
$result['dashboard content: '][] = array('news_id' => $news_id, 'image_url' => $image_url, 'news_title' => $news_title, 'news_description' => $news_description, 'news_article' => $news_article);
echo json_encode($result);
}
// do {
// $news_id = $rows['news_id'];
// $image_url = $rows['image_url'];
// $news_title = $rows['news_title'];
// $news_description = $rows['news_description'];
// $news_article = $rows['news_article'];
// $result['dashboard content: '][] = array('news_id' => $news_id, 'image_url' => $image_url, 'news_title' => $news_title, 'news_description' => $news_description, 'news_article' => $news_article);
// echo json_encode($result);
// } while ($rows = mysqli_fetch_assoc($statement));
mysqli_free_result($statement);
}
}
?>
推荐答案
这应该有效.您将要使用do...while
语句,否则将跳过第一个结果.
This should work. You'll want to use the do...while
statement otherwise the first result is skipped.
<?php
$dashboard_content_token = $_REQUEST["dashboard_content_token"];
$token = "g4";
require(cc_scripts/connect.php);
$sql = "SELECT * FROM `dashboard_content`";
$check = strcmp("$token", "$dashboard_content_token");
$statement = mysqli_query($con, $sql);
if (check) {
$rows = mysqli_fetch_assoc($statement);
if (!$rows) {
echo "No results!";
} else {
do {
$news_id = $rows['news_id'];
$image_url = $rows['image_url'];
$news_title = $rows['news_title'];
$news_description = $rows['news_description'];
$news_article = $rows['news_article'];
$result['dashboard content: '][] = array('news_id' => $news_id, 'image_url' => $image_url, 'news_title' => $news_title, 'news_description' => $news_description, 'news_article' => $news_article);
} while ($rows = mysqli_fetch_assoc($statement));
mysqli_free_result($statement);
echo json_encode($result);
}
}
?>
关键是将所有结果放入一个数组,然后只执行一个json_encode()
.当您多次调用json_encode()
时,您的API将返回无效的json.
The key is to put all of you results into an array and then just do one json_encode()
. When you call json_encode()
multiple times, your API will return invalid json.
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