php - 检查 mysql 中的表的名称

Question

我正在尝试实现一种方法来防止在 Android 应用程序中输入并发送到服务器的名称再次被使用。我认为最简单的方法是创建另一个表，每次添加产品时，名称都会添加到名称表中。我对 php 很陌生，所以这似乎是一个非常简单的问题，但我将如何检查表以查看名称是否已在其中。

这是我到目前为止所做的(其中大部分内容已经存在，只是注释掉了我的思考过程)

<?php
// array for JSON response
$response = array();
// check for required fields
if (isset($_POST['name']) && isset($_POST['longitude']) && isset($_POST['latitude']) && isset($_POST['pavement']) && isset($_POST['traffic']) && isset($_POST['environment'])) {
//if(Name is not already in list of names){
$name = $_POST['name'];
$longitude = $_POST['longitude'];
$latitude = $_POST['latitude'];
$pavement = $_POST['pavement'];
$traffic = $_POST['traffic'];
$environment = $_POST['environment'];

// include db connect class
require_once __DIR__ . '/db_connect.php';

// connecting to db
$db = new DB_CONNECT();

// mysql inserting a new row
$result = mysql_query("INSERT INTO spots(name, longitude, latitude, pavement, traffic,     environment) VALUES('$name', '$longitude', '$latitude', '$pavement', '$traffic', '$environment')");
//add new name to table
//$result2 = mysql_query("INSERT INTO names(name) VALUES('$name')");
// check if row inserted or not
if ($result) {
    // successfully inserted into database
    $response["success"] = 1;
    $response["message"] = "Spot successfully created.";

    // echoing JSON response
    echo json_encode($response);
} else {
    // failed to insert row
    $response["success"] = 0;
    $response["message"] = "Oops! An error occurred.";

    // echoing JSON response
    echo json_encode($response);
}
    /*
    } else {
    // name already taken
    $response["success"] = 0;
    $response["message"] = "Name has already been taken.";

    // echoing JSON response
    echo json_encode($response);
}
    */
  } else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";

// echoing JSON response
echo json_encode($response);
}
?>

很抱歉问了一个简单的问题，但提前谢谢您，

泰勒

Answer 1

您可以执行如下所示的 SELECT 查询:

$query = mysql_query("SELECT * FROM spots WHERE name='$name'");
$count = mysql_num_rows($query);
if($count == 0){
    // name not in database
}
else{
    // name is in database
}

但是，这使用了已弃用的mysql_函数。请使用mysqli_函数代替。