如何在更改时在php中创建“条件if”？ [重复]

Question

这个问题在这里已有答案：

• What is the difference between client-side and server-side programming? 5个答案

视图

<input type="text" id="textToDisplay" />
<br />
<input type="button" value="Add message" onClick="showMessage()" />
<br />
<span id="messageSpan"><span>

JS

function showMessage() {
  var message = jQuery("#textToDisplay").val();
  jQuery("#messageSpan").text(message);
}

顺便说一句，我正在使用laravel。基本上，我想要做的是，如果用户输入22，那么它将显示成功消息。所有这些都在PHP中进行，而无需更改/添加js。 jsfiddle example

基本上我的想法是这样的，

<?php 
if(<span id="messageSpan"><span>  == 22)
    success
else
    <span id="messageSpan"><span>
?>

Answer 1

你不能像上面描述它那样做。您要么只使用javascript / jquery，要么使用这两种语言（php和javascript）并向服务器发送ajax请求并在其中执行某些操作并将一些数据返回给客户端（javascript）并在您的视图中显示。

我添加了一些评论，以便更好地理解这个过程，如果您有任何问题请问。

我会使用ajax这样做，所以你将使用javascript和php，这对你来说应该更好。

所以这是你的HTML标记：

<input type="text" id="textToDisplay" />
<br />
<input type="button" value="Add message" onClick="showMessage()" />
<br />
<span id="messageSpan"><span>

现在，您将使用ajax将输入值发送到控制器，如下所示：

$(document).ready(function(){

    // on each key up we are going to sent the value later you should probably change this and sent the value after you click on a button
    $('#sentMessage').keyup(function(){

        // this holds the current value of the input field
        var inputValue = $(this).val();

        // you are going to sent the input data to this route '/showMessage'
        // you are going to use the post method
        $.ajax({
            type: "POST",
            url: '/showMessage',
            data: {
                inputValue: inputValue
            },
            // if you post is successful then do following ....
            success: function(data) {
                // here we are just going to show the message inside you span, the returned data comes from your controller
                $("#messageSpan").text(data.inputValue);
            },
            // if there was an error do following ....
            error: function(data) {
                // if there was an error you could just show a message with some error message
                $("#messageSpan").text('ERROR!');
            }
        });

    });

});

在Laravel Controller类中，您需要具有如下函数：

public function showMessage(Request $request)
{
        // $inputValue contains the inputValue data from your ajax call
        $inputValue = $request->inputValue;

        // You could / should  also do some validation here later on ...
        // Thus you should probably save the data, I do not know what excatly you are going to do with it later ...

        // now we are sending back the data
        return response()->json($inputValue);
});

你还需要做些什么来创建一个合适的路由，以便你的ajax请求找到它。所以在你的routes.php中你需要创建一个这样的路由：

Route::get('/showMessage', ['as' => 'showMessage', 'uses' => 'YourController@showMessage']);