Question

我有一个index.php和output.php 我需要两个按钮链接到output.php但不同的功能，我不能让它工作。这是我的代码：

在index.php中：

<?php 
    include 'output.php';
    if (isset ($_POST ['choice3'])){
        choice3();
    }
    if (isset ($_POST ['choice4'])){
        choice4();
    }
 ?>
<form method="post" action="output.php">
    <input type="submit" name="choice3" id="choice3" value="Choice 3">
    <input type="submit" name="choice4" id="choice4" value="Choice 4">      
</form>

在Output.php中，

function choice3(){} and function choice4(){}

每个功能都有不同的输出。但现在，当我点击按钮时，没有任何显示。

感谢您的帮助!!

Answer 1

只需用此

替换您的代码即可

<?php
include 'output.php';
$choice = $_POST['choice'];
if ($choice == 'choice3') {
    choice3();
} else if ($choice == 'choice4') {
    choice4();
}
?>
<script>
    function make_choice(choice) {
        document.getElementById('choice').value = choice;
        document.choice_form.submit();
    }
</script>
<form method="post" action="" name="choice_form">
    <input type="hidden" name="choice" id="choice" value="" />
    <input type="button" name="choice3" id="choice3" value="choice3" onclick="make_choice(this.value);">
    <input type="button" name="choice4" id="choice4" value="choice4" onclick="make_choice(this.value);">
</form>

Answer 2

将index.html.php和output.php替换为

<强>的index.php

<form method="post" action="output.php">
    <input type="submit" name="choice3" id="choice3" value="Choice 3">
    <input type="submit" name="choice4" id="choice4" value="Choice 4">      
</form>

<强> Output.php

<?php

if (isset ($_POST ['choice3']))
{
/ body of the choice 3 function without the function signature
}
elseif(isset ($_POST ['choice4']))
{
/ body of the choice 4 function without the function signature
}
    ?>

链接到PHP函数

2 个答案: