Question

您好我正在尝试创建一个小的电子邮件订阅者脚本，用于向我的数据库添加电子邮件地址。我能够使用像mysql_connect这样的弃用函数使其工作正常但是使用mysqli更好。但由于某种原因，我无法再将电子邮件地址插入数据库，但我可以正常连接到数据库，并检查电子邮件地址是否已经存在。我想知道为什么我的INSERT似乎不起作用。我没有多少PHP使用经验。

    if (!$link) {
    echo "save_failed cannot connect";   //if cant connect show error
    return; 
    }
    mysqli_select_db($link,$mydb);

// Clean variables before performing insert
$clean_email = mysqli_real_escape_string($link,$_POST['email']);
$clean_subscriber = mysqli_real_escape_string($link,$_POST['firstname']);
$clean_date = mysqli_real_escape_string($link,$_POST['date']);

$query = "SELECT * FROM EmailList WHERE email = '{$email}'";//check if already in list
$result = mysqli_query($link,$query);
$row_cnt = mysqli_num_rows($result);


if($row_cnt == 0) {   
// Perform insert if not in list
$sql = "INSERT INTO EmailList (Email,Name,Date) VALUES   
 ('$clean_email','$clean_subscriber','$clean_date')";
     echo "Thank you for Subscribing to my blog!";
} else {
    echo "You have subscribed already.  Thank you for subscribing";
}

Answer 1

看起来你错过了运行查询的行：

mysqli_query($link,$sql);

Answer 2

您需要使用

执行查询

mysqli_query（）

像这样：

    if (!$link) {
    echo "save_failed cannot connect";   //if cant connect show error
    return; 
    }
    mysqli_select_db($link,$mydb);

// Clean variables before performing insert
$clean_email = mysqli_real_escape_string($link,$_POST['email']);
$clean_subscriber = mysqli_real_escape_string($link,$_POST['firstname']);
$clean_date = mysqli_real_escape_string($link,$_POST['date']);

$query = "SELECT * FROM EmailList WHERE email = '{$email}'";//check if already in list
$result = mysqli_query($link,$query);
$row_cnt = mysqli_num_rows($result);


if($row_cnt == 0) {   
// Perform insert if not in list
$sql = "INSERT INTO EmailList (Email,Name,Date) VALUES   
 ('$clean_email','$clean_subscriber','$clean_date')";
 mysqli_query($link,$sql)
     echo "Thank you for Subscribing to my blog!";
} else {
    echo "You have subscribed already.  Thank you for subscribing";
}

php mysql连接数据库，读取但不插入

2 个答案: