Question

启动一个新项目，我已经从弃用的mysql_query命令转移到新的mysqli类型参数。

代码如下：

<?php

$mysqli = new mysqli("hostname", "username", "password", "dbname");

// Check the connection

if (mysqli_connect_errno()) {
    printf("Connect failed.");
    exit();
}

$sql = $mysqli->query("SELECT * FROM SafariFinder_Directory");

while ($rows = $sql->fetch_assoc()) {
    echo json_encode($rows);
}

echo $res;

$mysqli->close();

?>

这会得到以下输出：

{"id":"3","inGameName":"Syrinathos","friendCode":"0000 0000 0000","safariType":"Normal","safariSlot1":"Aipom","safariSlot2":"Kecleon","safariSlot3":""}
{"id":"2","inGameName":"Herschel","friendCode":"1234 12341 234","safariType":"Zombie","safariSlot1":"Crawler","safariSlot2":"Runner","safariSlot3":""}
{"id":"4","inGameName":"Syrinathos","friendCode":"0000 0000 0000","safariType":"Normal","safariSlot1":"Aipom","safariSlot2":"Kecleon","safariSlot3":"Ditto"}
{"id":"6","inGameName":"Kira","friendCode":"2345 2345 2345","safariType":"Ghost","safariSlot1":"Shuppet","safariSlot2":"Haunter","safariSlot3":"Gengar"}

缺少有效JSON的唯一组件是方括号和实际条目之间的逗号。非常感谢任何帮助。

Answer 1

您应该将结果放入数组中，然后对数组进行编码。

$rows = array();
while ($row = $sql->fetch_assoc()) {
    $rows[] = $row;
}

echo json_encode($rows);

无效的json_encode输出

1 个答案: