如何將單選按鈕的值發佈到MySQL

我一直在努力解決我的問題。所以在這裏，我正在構建一個表單，通過我的管理面板將新的在線遊戲插入到我的網站，一切工作都很完美，但後來我添加了遊戲類別，所以當我上傳遊戲時，我可以選擇哪個類別的頁面被上傳到... 因此，我爲每個類別選擇了一個單選按鈕，並且試圖使用$ _POST方法將特定類別的值帶到MySQL中，並且出於某種原因，遊戲無需上傳到MySQL在「game_category」空值...如何將單選按鈕的值發佈到MySQL

這裏是我的代碼：

<!DOCTYPE HTML> 
<html> 
<head> 
</head> 
<body> 

<form method="post" id="insert_form" action="insert_game.php" enctype="multipart/form-data"> 

<div class="insert_form_title"> 
<h1>Insert New Game Here</h1> 
</div> 

<div class="insert_form_inline"> 
<label class="insert_form_field" for="game_name">Game name:</label> 
<input type="text" name="game_name"> 
</div> 

<div class="insert_form_inline"> 
<label class="insert_form_field" for="game_category">Game category:</label> 
<input class="radio" type="radio" name="game_category" value="action"<?php print $action_status; ?>/> <span>Action</span> 
<input class="radio" type="radio" name="game_category" value="sports"<?php print $sports_status; ?>/> <span>Sports</span> 
</div> 

<div class="insert_form_inline"> 
<label class="insert_form_field" for="game_keywords">Game keywords:</label> 
<textarea name="game_keywords" cols="60" rows="15"></textarea> 
</div> 

<div class="insert_form_inline"> 
<label class="insert_form_field" for="game_image">Game image:</label> 
<input type="file" name="game_image"> 
</div> 

<div class="insert_form_inline"> 
<label class="insert_form_field" for="game_code">Game code:</label> 
<input type="file" name="game_code"> 
</div> 

<div class="insert_form_inline"> 
<label class="insert_form_field" for="game_file">Game flash file:</label> 
<input type="file" name="game_file"> 
</div> 

<div class="insert_form_inline"> 
<label class="insert_form_field" for="game_desc">Game description:</label> 
<textarea name="game_desc" cols="60" rows="15"></textarea> 
</div> 

<div class="submit"> 
<input type="submit" name="submit" value="Publish Game Now"></td> 
</div> 

</form> 

</body> 
</html> 
<?php 
include("../includes/connect.php"); 

$action_status = 'unchecked'; 
$sports_status = 'unchecked'; 

if(isset($_POST['submit'])){ 

$game_name = $_POST['game_name']; 
$game_category = $_POST['game_category']; 
$game_keywords = $_POST['game_keywords']; 
$game_image = $_FILES['game_image']['name']; 
$image_tmp = $_FILES['game_image']['tmp_name']; 
$game_code = $_FILES['game_code']['name']; 
$code_tmp = $_FILES['game_code']['tmp_name']; 
$game_file = $_FILES['game_file']['name']; 
$file_tmp = $_FILES['game_file']['tmp_name']; 
$game_desc = $_POST['game_desc']; 

if($game_name=='' or $game_category='' or $game_keywords=='' or $game_image=='' or $game_code=='' or $game_file=='' or $game_desc==''){ 

echo "<script>alert('Please enter all the fields below!')</script>"; 

exit(); 

} 
else { 

$path = "../games/games_files/$game_name"; 

mkdir("$path", 0777); 

move_uploaded_file($image_tmp,"../images/games_images/$game_image"); 

move_uploaded_file($code_tmp,"$path/$game_code"); 

move_uploaded_file($file_tmp,"$path/$game_file"); 

$insert_query = "insert into games (game_name,game_category,game_keywords,game_image,game_code,game_file,game_desc) values ('$game_name','$game_category','$game_keywords','$game_image','$game_code','$game_file','$game_desc')"; 

if($game_category == 'action'){ 

$action_status = 'checked'; 

}else if($game_category == 'sports'){ 

$sports_status = 'checked'; 

}else if(mysql_query($insert_query)){ 

echo "<script>alert('The Game Uploaded Successfully!')</script>"; 

echo "<script>window.open('view_games.php','_self')</script>"; 

} 

} 

} 

?> 

<?php } ?>

幫助的人？ :(

來源

2014-02-13 davidgpilot

我不確定什麼值=「動作」<？php print $ action_status; ？>你的代碼的一部分，它可能會導致錯誤，因爲該值應該只是「行動」編輯：我只是滾動，看到未選中。好吧 – Chitowns24

那麼我的代碼還有什麼問題呢？：o – davidgpilot

我不知道這個腳本將如何基於它被寫入的方式工作。我建議你把所有的PHP相關的代碼上面的HTML代碼 – Lepanto

大衛嘗試更新腳本如下。

$insert_query = "insert into games (game_name,game_category,game_keywords,game_image,game_code,game_file,game_desc) values ('$game_name','$game_category','$game_keywords','$game_image','$game_code','$game_file','$game_desc')"; 
if($game_category == 'action'){ 
    $action_status = 'checked'; 
}else if($game_category == 'sports'){ 
    $sports_status = 'checked'; 
} 

if(mysql_query($insert_query)){ 
    echo "<script>alert('The Game Uploaded Successfully!')</script>"; 
    echo "<script>window.open('view_games.php','_self')</script>"; 
}

而且也，在大if條件，你可以看到你在$ game_category = ''錯過了一個=跡象。

所以正確下面

if($game_name=='' or $game_category='' or $game_keywords=='' or $game_image=='' or $game_code=='' or $game_file=='' or $game_desc==''){

到

if($game_name=='' or $game_category=='' or $game_keywords=='' or $game_image=='' or $game_code=='' or $game_file=='' or $game_desc==''){

來源

2014-02-13 08:52:28 Lepanto

謝謝！） – davidgpilot

儘管如此，我在phpMyAdmin的「game_category」下沒有獲得任何價值...... :( – davidgpilot

你可以把'echo $ insert_query; exit;'before'if（$ game_category = ='action'）{'然後看看你是否獲得了game_category的價值？ – Lepanto

該解決方案如下： 1.創建一個「if」語句如下面的代碼：

if($game_category == '3d'){ 

$three_d_status = 'checked'; 

}else if($game_category == 'action'){ 

$action_status = 'checked'; 

}else if($game_category == 'adventure'){ 

$adventure_status = 'checked'; 

}

2.然後，創建您想使用的每個領域一個「單選按鈕」這樣一行：

<input class="radio" type="radio" name="game_category" value="3d"<?php print $three_d_status; ?>/><span>3D</span>

就是這樣，你準備好了:)

來源

2014-02-16 13:44:27 davidgpilot

如何將單選按鈕的值發佈到MySQL

回答

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