如何在不提交的情況下獲取表單之外的輸入控件的值？

Question

我花了將近一個小時，谷歌搜索，但我沒有發現任何有用的， 這裏是我的代碼的形式外

<div class = "form-group col-md-12"> 
    <input type="text" name="filename" id="filename"></input> 
</div> 

它，我想要做的是什麼我輸入這裏應該把

$fopen=fopen($uploadpath.here.'.png','wb'); 

所以我可以管理文件名，因此它不會是靜態的。

Answer 1

您必須將數據發送到您的後端代碼（php），因爲它一旦涉及到瀏覽器就不會運行。如果我是對的，你想在後臺打開文件，但你的頁面可能會重新加載，所以你不想提交表單。如果是這種情況下，你可以使用Ajax，就像這樣：

<div class = "form-group col-md-12"> 
    <input type ="text" name = "filename" id="filename"></input> 
</div> 
<script> 
var filename = $("#filename").val(); 
$.ajax({ 
    url: myfile.php, //url to your php file with code $fopen=fopen($uploadpath.here.'.png','wb'); 
    type: post, 
    data: {"fileName":filename}, 
    success: function(result) { 
     // use your code to handle the result whatever you want to do here 
    } 
}) 
</script> 

你的PHP文件

myfile.php

<?php 
$uploadPath = $_POST['fileName']; 
$fopen=fopen($uploadpath.here.'.png','wb'); 
echo "success"; // you can return whatever you want and that will go to success function in your javascript 

?> 

Answer 2

使用jQuery AJAX $。員額它更容易

<div class = "form-group col-md-12"> 
    <input type ="text" name = "filename" id="filename"></input> 
</div> 
<script> 
var filename = $("#filename").val(); 
$.post("yourfile.php", {"fileName":filename}, function(data) { 
    // do something 
}) 
</script> 

您的php

<?php 
    $uploadPath = 'youruploadfolder'; // this is the location of your file folder 
    $fileName = $_POST['fileName']; // this is your file passed via ajax request 
    $fopen = fopen($uploadpath.$fileName.'.png','wb'); 
    echo "success"; // you can return whatever you want and that will go to success function in your javascript 
?>