Question

在我的网络服务中，我有一个用于插入经度和纬度值的页面。成功插入数据库后，将在表中创建一个新行，并使用键＆＃34; id＆＃34; （INT）是自动增量。

它返回一个JSON对象：

{＆＃34;成功＆＃34;：1，＆＃34;消息＆＃34;：＆＃34;现货成功添加！＆＃34;}

我的问题：我想将为每行创建的id添加到生成的JSON对象中。像这样：

{＆＃34;成功＆＃34;：1，＆＃34;消息＆＃34;：＆＃34;现货成功添加！＆＃34;，＆＃34; id＆＃34;：54}

这是我到目前为止所得到的。行被插入到表中，但是我无法在JSON对象中显示我的id：

{＆＃34;成功＆＃34;：1，＆＃34;消息＆＃34;：＆＃34;现货成功添加！＆＃34;，＆＃34; like_id＆＃34;：null}

idtest.php

<?php
//load and connect to MySQL database stuff
require("config.inc.php");

if (!empty($_POST)) {


    //initial query
    if(isset($_POST['user_id']) && !empty($_POST['user_id'])) {
        $query = "INSERT INTO spot ( latitude, longitude, user_id ) VALUES ( :lat, :long, :uid ) ";
        //Update query
        $query_params = array(
            ':lat' => $_POST['latitude'],
            ':long' => $_POST['longitude'],
            ':uid' => $_POST['user_id']
            );

    } else {
        $query = "INSERT INTO spot ( latitude, longitude ) VALUES ( :lat, :long ) ";
        //Update query
        $query_params = array(
            ':lat' => $_POST['latitude'],
            ':long' => $_POST['longitude']
            );
    }
    //execute query
    try {
        $stmt   = $db->prepare($query);
        $result = $stmt->execute($query_params);
    }
    catch (PDOException $ex) {
        // For testing, you could use a die and message. 
        //die("Failed to run query: " . $ex->getMessage());

        //or just use this use this one:
        $response["success"] = 0;
        $response["message"] = "Database Error. Couldn't add lat/long-pair!";
        die(json_encode($response));
    }

    $latt = $_POST['latitude'];
    $longg = $_POST['longitude'];

    $getId = "SELECT id FROM spot WHERE latitude=$latt AND longitude=$longg LIMIT 1";
    echo $getId;
    //execute query
    try {
        $stmt2   = $db->prepare($getId);
        $result2 = $stmt2->execute($query_params);
    }
    catch (PDOException $ex) {
        // For testing, you could use a die and message. 
        //die("Failed to run query: " . $ex->getMessage());

        //or just use this use this one:
        $response["success"] = 0;
        $response["message"] = "Database Error. Couldn't retrieve like_id!";
        die(json_encode($response));
    }

    $row = $stmt2->fetchAll();
    if($row) {

    $response["success"] = 1;
    $response["message"] = "Spot Successfully Added!";
    $response["like_id"] = $row["id"];
    echo json_encode($response);
    }

} else {
    ?>
    <h1>Registrer GPS-koordinater</h1>
    <form action="idtest.php" method="post">
        Latitude:<br />
        <input type="text" name="latitude" value="" />
        <br /><br />
        Longitude:<br />
        <input type="text" name="longitude" value="" />
        <br /><br />
        Google ID:<br />
        <input type="text" name="user_id" value="" />
        <br /><br />
        <input type="submit" value="Opprett kolonne i 'spot'" />
    </form>
    <?php
}

?>

Answer 1

更改此行

if($row) {

到

if(!$row) {

因为在此之前找不到$ row

在同一个PHP文件中同时执行INSERT和SELECT

1 个答案: