php - 意外的错误 -

Question

我对 PHP 很陌生，我只是在看一个设置 WAMP 服务器并使用 PHP 与服务器连接的示例。我将一些数据插入数据库的表中，并希望使用 PHP 文件检索所有数据。

当我尝试通过这样做检查输出时:

“localhost/android_connect/get_all_products.php”

我收到此错误:

解析错误:语法错误，第 5 行 C:\wamp\www\android_connect\get_all_products.php 中出现意外的“$response”(T_VARIABLE)

这是示例:

    <?php
    // array for JSON response

    $response = array();
    $db = new DB_CONNECT(); 
    $result = mysql_query("SELECT *FROM products") or die(mysql_error());

    if (mysql_num_rows($result) > 0) {
        $response["products"] = array();
        while ($row = mysql_fetch_array($result)) {

            $product = array();
            $product["id"] = $row["id"];
            $product["name"] = $row["name"];
            $product["price"] = $row["price"];
            $product["description"] = $row["description"];
            $product["created_at"] = $row["created_at"];
            $product["updated_at"] = $row["updated_at"]; 
            array_push($response["products"], $product);

        }
        
        $response["success"] = 1;
        echo json_encode($response);
     } else {

        $response["success"] = 0;
        $response["message"] = "No products found";
        echo json_encode($response);

    }
    ?>

我知道它的语法错误，但我找不到确切的位置。

Answer 1

我已经修改了您的代码，请检查下面

<?php
// array for JSON response

$response = array();
$db = new DB_CONNECT(); 
$result = mysql_query("SELECT *FROM products") or die(mysql_error());

if (mysql_num_rows($result) > 0) {
    while ($row = mysql_fetch_array($result)) {

        $product = array();
        $product["id"] = $row["id"];
        $product["name"] = $row["name"];
        $product["price"] = $row["price"];
        $product["description"] = $row["description"];
        $product["created_at"] = $row["created_at"];
        $product["updated_at"] = $row["updated_at"]; 
        $products[] = $product;

    }

    $response["products"] = $products;
    $response["success"] = 1;
    echo json_encode($response);
 } else {

    $response["success"] = 0;
    $response["message"] = "No products found";
    echo json_encode($response);

}
?>