php - 如何通过 php 从 mysql 数据库返回多行/所有行并制作 json？

Question

我的 php 文件中有这个:

$response = array();
    $user = $db->test($email);
                if ($user) {
                    // user stored successfully
                    $response["success"] = 1;
                    $response["user"]["email"] = $user["email"];
                    $response["user"]["imagepath"] = $user["imagepath"];
                    $response["user"]["about"] = $user["about"];

                    echo json_encode($response);
                } else {
                    // user failed to store
                    $response["error"] = 1;
                    $response["error_msg"] = "JSON Error occured in db";
                    echo json_encode($response);
                }

测试函数是

public function test($email){
            $result = mysql_query("SELECT * FROM activities WHERE email = '$email'");
            // return user details
            return mysql_fetch_array($result);
    }

测试函数返回多行，但 php 仅发送数据库中与该电子邮件匹配的第一行的 json。如何制作 foreach() 或类似的东西来对测试函数返回的所有内容进行编码？

Answer 1

首先你必须拒绝使用 mysql_fetch_array() 函数(读 this )。

接下来，您必须了解变量“user”包含所有找到的用户的列表(公平地说，它应该称为“users”)

$response = array();
$users = $db->test($email);
foreach($users as $user){
    if ($user) {
        // user stored successfully
        $response["success"] = 1;   
        $response["users"][] = array(
            "email" => $user["email"],
            "imagepath" => $user["imagepath"],
            "about" => $user["about"]
        );
    } else {
        if( empty($response) ){
            // user failed to store
            $response["error"] = 1;
            $response["error_msg"] = "JSON Error occured in db";
        }
    }
}
echo json_encode($response);

并更新测试功能:

public function test($email){
    $result = mysql_query("SELECT * FROM activities WHERE email = '$email'");
    // get all users details
    $users = array();
    while($user = mysql_fetch_array($result)){
        $users[] = $user;
    }
    return $users;
}