如何正确使用PHP将MySQL对象编码为JSON? [英] How do I properly use PHP to encode MySQL object into JSON?

问题描述

我正在尝试遍历MySQL对象，并在另一页上使用ajax调用来追加数据，但我无法获取php将有效的JSON返回给回调.

这显然不起作用...

<?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);

?>

或者这一个...

<?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 );

这应该做到.另外，您可以使用 http://jsonlint.com/来查看JSON输出有什么问题.

更新:使用fetch_all()也是一个好主意

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

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