jquery - 確認對話框 - 工作php代碼

Question

我有一個關於jquery的問題。我的代碼如下所示：

<button id="callConfirm'.$row['id'].'">DEL</button> 


    <div id="dialog" title="Realy?"> 
Realy delete '.$row['marks'] .' s váhou '.$row['weight'] .' z '.$row['id'] .' ? 
</div>​ 

<script type="text/javascript"> 
$("#dialog").dialog({ 
autoOpen: false, 
modal: true, 
buttons : { 
    "Opravdu" : function() { 

    alert("You have confirmed!");  }, 
    "Ne" : function() { 
     $(this).dialog("close"); 
    } 
    } 
}); 

$("#callConfirm'.$row['id'].'").on("click", function(e) { 
e.preventDefault(); 
$("#dialog").dialog("open"); 
}); 
</script>' 

，我需要爲了點擊到Opravdu執行delele.php腳本，看起來像這樣：

<?php 
include 'conn.php'; 
$back = urldecode($_GET['back']); 
$id = mysql_real_escape_string($_GET['id'] * 1); 
mysql_query("DELETE FROM znamky WHERE id=$id"); 
?> 

Answer 1

"Opravdu" : function() { 
    $.get("delete.php?id=...&back=...",function(data) { 
    alert(data); 
    }); 
} 

，其中數據從刪除返回

以下是您可能想要如何將ID傳遞給gialog

jquery-ui, Use dialog('open') and pass a variable to the DIALOG