javascript 如何从 mysqli 结果构建正确的 json？
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How can I build a correct json from mysqli result?
提问by prog_24
I am trying to build a json object from my mysqli result. How do I go about it. At the moment it does not create a json looking object.
我正在尝试从我的 mysqli 结果构建一个 json 对象。我该怎么做。目前它不会创建一个看起来像 json 的对象。
Here is my code:
这是我的代码：
$result = $dataConnection->prepare("SELECT id, artist, COUNT(artist) AS cnt FROM {$databasePrefix}users GROUP BY artist ORDER BY cnt DESC LIMIT 0 , 30");
$result->execute();
if($result->error)
{
die("That didn't work. I get this: " . $result->error);
}
$result->bind_result($id, $artist, $count);
$data = array();
while($result->fetch()){
$data[] = '{ id :'.$id.', artist :'.$artist.', count :'.$count.'}';
}
echo json_encode($data);
$dataConnection->close();
I want a data object like:
我想要一个数据对象，如：
{"id":"27","artist":"myArtist","count":"29"},....
回答by Your Common Sense
$result = $dataConnection->query("SELECT id, artist, COUNT(artist) AS count FROM {$databasePrefix}users GROUP BY artist ORDER BY cnt DESC LIMIT 0 , 30");
$data = array();
while($row = $result->fetch_assoc()){
$data[] = $row;
}
echo json_encode($data);
To tell you truth, mysqli is awful API to be used right in the application code.
说实话，mysqli 是在应用程序代码中使用的糟糕的 API。
Do yourself a favor and use at least PDO
帮自己一个忙，至少使用PDO
$result = $dataConnection->prepare("SELECT id, artist, COUNT(artist) AS count FROM {$databasePrefix}users GROUP BY artist ORDER BY cnt DESC LIMIT 0 , 30");
$result->execute();
echo json_encode($result->fetchAll());
unlike mysqli, it's methods always works.
与 mysqli 不同，它的方法总是有效的。
回答by Orangepill
Don't build your json for the values array that you will call json_encode on
不要为将在其上调用 json_encode 的 values 数组构建 json
instead of:
代替：
$data[] = '{ id :'.$id.', artist :'.$artist.', count :'.$count.'}';
do
做
$data[] = array("id"=>$id, "artist"=>$artist, "count"=>$count);
回答by BenRoe
If you use mysqli here is an example. I use it in combination with a javascript ajax call.
The output looks like this:
[{"field1":"23","field2":"abc"},{"field1":"24","field2":"xyz"}]
如果您使用 mysqli，这里是一个示例。我将它与 javascript ajax 调用结合使用。
输出如下所示：
[{"field1":"23","field2":"abc"},{"field1":"24","field2":"xyz"}]
$mysqli = mysqli_connect('localhost','dbUser','dbPassword','dbName');
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "SELECT field FROM table LIMIT 10";
if ($result = mysqli_query($mysqli, $query)) {
$out = array();
while ($row = $result->fetch_assoc()) {
$out[] = $row;
}
/* encode array as json and output it for the ajax script*/
echo json_encode($out);
/* free result set */
mysqli_free_result($result);
/* close connection*/
$mysqli->close();
}
/* close connection*/
$mysqli->close();
回答by Populus
just create an array of all your rows, then do:
只需创建一个包含所有行的数组，然后执行以下操作：
echo json_encode($array)
