如何正确使用 PHP 将 MySQL 对象编码为 JSON?

我正在尝试遍历 MySQL 对象并在另一个页面上使用 ajax 调用来附加数据，但我无法让 php 向回调返回有效的 JSON.

这个显然行不通...

query($myQuery) or die($mysqli->error);$row = $result->fetch_assoc();回声 json_encode($row);?>

或者这个...

query($myQuery) or die($mysqli->error);while ( $row = $result->fetch_assoc() ){回声 json_encode($row) .", ";}?>

解决方案

$data = array();while ( $row = $result->fetch_assoc() ){$data[] = json_encode($row);}回声 json_encode( $data );

这应该可以.此外，您可以使用 http://jsonlint.com/ 查看您的 JSON 输出有什么问题.>

更新:使用 fetch_all() 也可能是个好主意

$data = $result->fetch_all( MYSQLI_ASSOC );回声 json_encode( $data );

I am trying to iterate through a MySQL object and use an ajax call on another page to append the data but I can't get the php to return valid JSON to the callback.

This one obviously doesn't work...

<?php

    $db_host = "localhost";
    $db_user = "blah";
    $db_pass = "blah";
    $db_name = "chat";
    $mysqli = new MySQLi($db_host, $db_user, $db_pass, $db_name);
    $myQuery = "SELECT * FROM users";
    $result = $mysqli->query($myQuery) or die($mysqli->error);
    $row = $result->fetch_assoc();
    echo json_encode($row);

?>

Or this one...

<?php

    $db_host = "localhost";
    $db_user = "blah";
    $db_pass = "blah";
    $db_name = "chat";
    $mysqli = new MySQLi($db_host, $db_user, $db_pass, $db_name);
    $myQuery = "SELECT * FROM users";
    $result = $mysqli->query($myQuery) or die($mysqli->error);
    while ( $row = $result->fetch_assoc() ){
        echo json_encode($row) . ", ";
    }

?>

解决方案

$data = array();

while ( $row = $result->fetch_assoc() ){
    $data[] = json_encode($row);
}
echo json_encode( $data );

This should do it. Also, you can use http://jsonlint.com/ to see what are the problems with your JSON output.

Update: using fetch_all() might be a good idea too

$data = $result->fetch_all( MYSQLI_ASSOC );
echo json_encode( $data );