Question

我想选择数据库更多的值（我已经这样做了）并转换为JSON

我尽力了

php

$a = $_GET['name'];

header('Content-Type: application/json');
echo '{"results":[';
$selectSearch = "SELECT * from `users` WHERE `name` LIKE '".$a["term"]."%'";
$rezultatul = $db->query($selectSearch);
if ($rezultatul->num_rows > 0) {
    while($row = $rezultatul->fetch_assoc()) {
        $name = $row["name"];

        $arr = array('id' => $row["id"], 'text' => $row["name"], 'level' => $row["Level"]);
        echo json_encode($arr);
    }
}


echo ']}';

他看起来像这样：

{"results":[{"id":"1","text":"Pompiliu","level":"7"}
{"id":"11","text":"Pompiliu1","level":"100"}]}

但是两者之间一定是这样

{"id":"1","text":"Pompiliu","level":"7"}, 
{"id":"11","text":"Pompiliu1","level":"100"}

什么时候会有3个结果

{"id":"1","text":"Pompiliu","level":"7"},
{"id":"11","text":"Pompiliu1","level":"100"},
{"id":"12","text":"Pompiliu2","level":"100"}

Answer 1

使用[]然后使用json_encode添加到数组。
不要尝试自己构建json字符串。

if ($rezultatul->num_rows > 0) {
    while($row = $rezultatul->fetch_assoc()) {
        $name = $row["name"];

        $arr[] = array('id' => $row["id"], 'text' => $row["name"], 'level' => $row["Level"]);
    }
}
echo json_encode(["results" => $arr]);

我对JSON PHP多值有疑问

1 个答案: