Question

我有一个包含多行的表，我想用json对所有行进行编码。

我一直在寻找其他问题和解决方案，我尝试了很多不同的方法，但json_encode仍然返回null

<?php
    $mysqli = new mysqli('localhost', 'root', 'password', 'testBasParmak');

    /* check connection */
    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }
    $sth = mysql_query("SELECT * FROM pictures");
    $rows = array();
    while($r = mysql_fetch_assoc($sth)) {
        $rows[] = $r;
    }
    print json_encode($rows); 

    $error = json_last_error();
    print  $error;

$mysqli->close();

?>

终端输出

[]0

如果我试试这个

$sth = mysql_query("SELECT * FROM pictures");
    $rows = array("id" => $id,"name" => $name,"description" => $description,"url" => $url,"users_id" => $users_id,"users_id" => $users_id,"totalvoteup" => $totalvoteup,"totalvotedown" => $totalvotedown,"totalvoteneutral" => $totalvoteneutral);
    while($r = mysql_fetch_assoc($sth)) {
        $row[] = $r;
    }
    print json_encode($rows); 

    $error = json_last_error();
    print  $error;

终端输出

{"id":null,"name":null,"description":null,"url":null,"users_id":null,"totalvoteup":null,"totalvotedown":null,"totalvoteneutral":null}0

也许它没有发送正确的查询？

Answer 1

$row[] = $r;

应该是

$rows[] = $r;

Answer 2

最后我发现答案我的数据库连接是mysqli，但我对SELECT *的查询是mysql所有我需要改变的是mysql命令到mysqli

$sth = $mysqli->query("SELECT * FROM pictures");
    $rows = array();
    while($r = mysqli_fetch_assoc($sth)) {
        $rows[]=$r;
    }
    print json_encode($rows);

给出正确的输出

["id":"2","name":"Jess","description":"who","url":"www","users_id":"705735067","totalvoteup":"0","totalvotedown":"0","totalvoteneutral":"0"},{"id":"3","name":"Jess","description":"who","url":"www","users_id":"705735067","totalvoteup":"0","totalvotedown":"0","totalvoteneutral":"0"},{"id":"4","name":"Jess","description":"who","url":"www","users_id":"705735067","totalvoteup":"0","totalvotedown":"0","totalvoteneutral":"0"}]

Json_encode返回Null Php / MYSQL

2 个答案: