Question

我有像这样的代码php

$result = mysql_query("SELECT phone FROM user where phone LIKE 'hjkhkjh');


// check if row inserted or not
if ($result) {
    // successfully inserted into database
    $response["success"] = 1;
    $response["message"] = "Account wasnt created.";

    // echoing JSON response
    echo json_encode($response);
}
else  {
    // failed to insert row
    $response["success"] = 0;
    $response["message"] = "Account was created.";

我想比较数据库中的电话号码，但是当我运行此代码时，此代码显示“帐户未创建”，而我的表中有数据电话号码'hjkhkjh'，我的代码的任何解决方案，谢谢

Answer 1

因为当if-construct解析为true时，你写的是“帐户未创建”。试试这个

if ($result) {
    // successfully inserted into database
    $response["success"] = 1;
    $response["message"] = "Account was created.";

    // echoing JSON response
    echo json_encode($response);
}
else  {
    // failed to insert row
    $response["success"] = 0;
    $response["message"] = "Account wasnt created.";

Answer 2

www.php.net/manual/en/function.mysql-query.php说：

对于SELECT，SHOW，DESCRIBE，EXPLAIN和其他语句返回 resultset，mysql_query（）在成功时返回资源，或者返回FALSE 错误。

因为您的查询不会产生错误，所以return是资源，它传递if ($result) { （无论数据库表中是否存在具有电话LIKE'hjkhkjh'的实际条目）

Answer 3

我找到了解决方案，我改变了我的代码 $ result = mysql_query（“来自用户的选择电话=电话='hjkhkjh'”）;

// check if row inserted or not
if($row = mysql_fetch_array($result)){
    // successfully inserted into database
    $response["success"] = 1;
    $response["message"] = "Account was created.";

    // echoing JSON response
    echo json_encode($response);
}
else  {
    // failed to insert row
    $response["success"] = 0;
    $response["message"] = "Account wasnt created.";

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

谢谢大家：D

比较数据sql

3 个答案: