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在 PHP 問題中插入數據庫中的按鈕和函數記錄

在 PHP 問題中插入數據庫中的按鈕和函數記錄

PHP
HUWWW 2023-05-26 17:32:21
這是代碼,不起作用 -<form method="post">    <input type="submit" name="zapis" id="zapisk" value="Запис" /><br/></form><?phpfunction zapisk(){$servername = "localhost";    $username = "test";    $password = "";    $dbname = "myDB";// Create connection    $conn = new mysqli($servername, $username, $password, $dbname);// Check connection    if ($conn->connect_error) {        die("Connection failed: " . $conn->connect_error);    }    $sql = "INSERT INTO MyGuests (firstname, lastname, email)VALUES ('John', 'Doe', '[email protected]')";    if ($conn->query($sql) === TRUE) {        echo "New record created successfully";    } else {        echo "Error: " . $sql . "<br>" . $conn->error;    }    $conn->close();}//{echo "Your test function on button click is working";}//if(array_key_exists('zapis',$_POST)){zapisk();}?>如果我激活最后兩行,此時作為評論,該功能將起作用。是否有可能,該功能可以在不使用最后兩行的情況下工作。
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1 回答

?
天涯盡頭無女友

TA貢獻1831條經驗 獲得超9個贊

zapisk提交表單時必須調用函數。


<form method="post">

    <input type="submit" name="zapis" id="zapisk" value="Запис" /><br/>

</form>


<?php


function zapisk(){

    $servername = "localhost";

    $username = "test";

    $password = "";

    $dbname = "myDB";


// Create connection

    $conn = new mysqli($servername, $username, $password, $dbname);

// Check connection

    if ($conn->connect_error) {

        die("Connection failed: " . $conn->connect_error);

    }


    $sql = "INSERT INTO MyGuests (firstname, lastname, email) VALUES ('John', 'Doe', '[email protected]')";


    if ($conn->query($sql) === TRUE) {

        echo "New record created successfully";

    } else {

        echo "Error: " . $sql . "<br>" . $conn->error;

    }


    $conn->close();

}


// Check form submitted and post has zapis value. 

if(!empty($_POST['zapis'])){

   zapisk();

}


?>



或者您不使用函數并在 if 語句中編寫代碼:


<form method="post">

    <input type="submit" name="zapis" id="zapisk" value="Запис" /><br/>

</form>


<?php


// Check form submitted and post has zapis value. 

if(!empty($_POST['zapis'])){

    $servername = "localhost";

    $username = "test";

    $password = "";

    $dbname = "myDB";


// Create connection

    $conn = new mysqli($servername, $username, $password, $dbname);

// Check connection

    if ($conn->connect_error) {

        die("Connection failed: " . $conn->connect_error);

    }


    $sql = "INSERT INTO MyGuests (firstname, lastname, email) VALUES ('John', 'Doe', '[email protected]')";


    if ($conn->query($sql) === TRUE) {

        echo "New record created successfully";

    } else {

        echo "Error: " . $sql . "<br>" . $conn->error;

    }


    $conn->close();


}


?>


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反對 回復 2023-05-26
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