2016-06-10 74 views
0

我有一個文本框和一個下拉菜單。這些顯示/隱藏兩個單選按鈕。 這個值=「hariciLink」值是張貼成功,但這個值=「dahiliLink」值是張貼空白。我該如何解決這個問題? 感謝您的答案。單選按鈕值=「dahiliLink」值空發佈

<script type="text/javascript"> 
$(document).ready(function(){ 
    $('input[type="radio"]').click(function(){ 
     if($(this).attr("value")=="hariciLink"){ 
      $(".box").not(".hariciLink").hide(); 
      $(".hariciLink").fadeIn(300); 
     } 
     if($(this).attr("value")=="dahiliLink"){ 
      $(".box").not(".dahiliLink").hide(); 
      $(".dahiliLink").fadeIn(300); 
     } 
    }); 
}); 
</script> 


<div class="form-group"> 
    <label for="ikon" class="col-lg-2 control-label">Bağlantı</label> 
    <div class="col-lg-10"> 
     <div class="radio"> 
     <label><input type="radio" name="link" value="hariciLink"> Harici Link</label> 
     </div> 
     <div class="radio"> 
      <label><input type="radio" name="link" value="dahiliLink"> Dahili Link</label> 
     </div> 
    </div> 
</div> 
<div class="box form-group dahiliLink"> 
    <label for="link" class="col-lg-2 control-label">Seçiniz</label> 
    <div class="col-lg-10 selectContainer"> 
     <select name="link" value="" class="form-control" required> 
      <option value="black">Black</option> 
      <option value="blue">Blue</option> 
      <option value="green">Green</option> 
      <option value="orange">Orange</option> 
      <option value="red">Red</option> 
      <option value="yellow">Yellow</option> 
      <option value="white">White</option> 
     </select> 
    </div> 
</div> 

<div class="box form-group hariciLink"> 
    <label for="link" class="col-lg-2 control-label">Link</label> 
    <div class="col-lg-10"> 
     <input type="text" name="link" placeholder="Örnek: http://www.websayfam.com" class="form-control"/> 
    </div> 
</div> 

$data = array(); 
if (isset($_GET['files'])) { 
    $error = false; 
    $files = array(); 
    $uniqid = uniqid(); 
    $uploaddir = '../../images/'.$uniqid; 
    foreach($_FILES as $file) { 
     if (move_uploaded_file($file['tmp_name'], $uploaddir.basename($file['name']))) { 
      $files[] = $uploaddir.$file['name']; 
     } else { 
      $error = true; 
     } 
    } 
    $data = ($error) ? array('error' = > 'There was an error uploading your files') : array('files' = > $files); 
} else { 
    $arr - > image = $file['name']; 
    $_SESSION['image'] = "img-".$file['name']; 
    $arr - > ok = "ok"; 
    $data = array('success' = > 'Form was submitted', 'formData' = > $file['name']); 
} 

回答

0

這是因爲當您有多個單選按鈕具有相同的名稱時,只有選定的單選按鈕的值作爲該名稱的值發佈。

+0

嗨,感謝您的回覆。我該如何解決它?因爲我有這樣的PHP表單。 $ isim = $ _POST ['isim']; $ ikon = $ _POST ['ikon']; $ ustmenude = isset($ _ POST ['ustmenudeGizle'])? $ _POST ['ustmenudeGizle']:0; $ link = $ _POST ['link'] == NULL? '#':$ _POST ['link']; – Dogan

+0

你想要發佈什麼?你想發送這兩個值到服務器?如果是這樣的話,你可以使用'input'標籤。如果您不想顯示這些輸入,則可以使用隱藏輸入'' –

+0

我不想發送這兩個值。只是我會選擇單選按鈕,然後我填充,然後我會發送。就是這樣。但如果我從下拉列表中選擇,則發佈空值。我該如何解決它? – Dogan