想從HTML獲得價值到PHP

Question

我當前的PHP代碼是：

<?php 
header('Content-type: image/jpeg'); 
$jpg_image = imagecreatefromjpeg('Desert.jpg'); 
$white = imagecolorallocate($jpg_image, 73, 41, 236); 
$font_path = 'OpenSans-Italic.TTF'; 
$text = $_GET['name'] ; 
imagettftext($jpg_image, 25, 0, 75, 50, $white, $font_path, $text); 
imagejpeg($jpg_image); 
imagedestroy($jpg_image); 
?> 

這是完全的工作我使用$_GET['name']獲得來自輸入標籤的用戶文字輸入在HTML中的名字是name="name"和現在我想從組合框中獲取值

我的組合框的html代碼：

<select name="gender" id="gender" class="form-control input-lg" maxlength="12"> 

> 
<option value="male">male</option> 
<option value="female">female</option> 

</select> 

當我試圖讓從$_GET['gender'];值圖像上未顯示用戶選擇的值。幫幫我！

我已經試過這也

<?php 
header('Content-type: image/jpeg'); 
$jpg_image = imagecreatefromjpeg('Desert.jpg'); 
$white = imagecolorallocate($jpg_image, 73, 41, 236); 
$font_path = 'OpenSans-Italic.TTF'; 
$text = $_GET['name'] ; 
$text2= $_GET['gender']; 
imagettftext($jpg_image, 25, 0, 75, 50, $white, $font_path, $text); 
imagettftext($jpg_image, 25, 0, 80, 60, $white, $font_path, $text2); 
imagejpeg($jpg_image); 
imagedestroy($jpg_image); 
?> 

但

Answer 1

您使用$text = $_GET['name'] ;不工作，但選擇具有name="gender"。您需要將選擇的名稱更改爲name或通過$_GET['gender']$_GET['gender']