爲什麼我的動態刪除不起作用？

插入和動態表創建工作正常。通過按下該行上的按鈕，所有事情都會按照從數據庫中刪除記錄的方式進行。嘗試了幾種不同的方式來訪問沒有結果的值。該按鈕只是點擊，沒有任何反應。任何幫助將不勝感激！（代號爲所有四個部分如下。）爲什麼我的動態刪除不起作用？

主頁：

<!doctype html> 
<html> 
<head> 
<meta charset="utf-8"> 
<title>PHP Test</title> 
<link href="styleTest.css" rel="stylesheet" type="text/css"> 
</head> 

<body> 
<div id="container"> 
<header id="header"><strong>PHP Test</strong></header> 
<div id="mainContent" align="center"> 
<form id="myForm" action="insert.php" method="post"> 
<table> 
    <tr valign="baseline"> 
     <td>Name:</td> 
     <td> <input id="name" type="text" name="name"/></td> 
    </tr> 
    <tr valign="baseline"> 
    <td>Comment:</td> 
     <td> <input id="comment" type="text" name="comment"/></td> 
    </tr> 
    <tr valign="baseline"> 
     <td><input name="sub" id="sub" type="button" value="Save"></td> 
    </tr> 
</table> 
</form> 
<span id="result"></span> 
</div><!-- mainContent --> 
</div><!-- Container --> 
<script src="jquery-1.8.1.min.js" type="text/javascript"></script> 
<script src="my_script.js" type="text/javascript"></script> 
</body> 
</html>

插入頁：

<?php 
//get the values passed by post 
$name = mysql_real_escape_string($_POST['name']); 
$comment = mysql_real_escape_string($_POST['comment']); 

//create connection to Database 
$hostname = "localhost"; 
$database = "test"; 
$username = "root"; 
$password = ""; 
$con = mysqli_connect($hostname, $username, $password) or trigger_error(mysql_error(),E_USER_ERROR); 
mysqli_select_db($con,$database); 

//Run the insert query and check to see if it processes 
if(mysqli_query($con,"INSERT INTO info VALUES('','$name', '$comment')")) 
{ 
    echo "Successfully Inserted"; 
    $sql2="SELECT * FROM info ORDER BY ID ASC"; 
    $result = mysqli_query($con,$sql2); 
    echo "<table border='1'> 
    <tr> 
    <th>Name</th> 
    <th>Comment</th> 
    <th>Remove Entry</th> 
    </tr>"; 

    while($row = mysqli_fetch_array($result)) 
    { 
    echo "<tr>"; 
    echo "<td>" . $row['NAME'] . "</td>"; 
    echo "<td>" . $row['COMMENTS'] . "</td>"; 
    echo "<td> <form id='myForm2' method='post' action='delete.php'><input name='del' id='del' 
    type='button' value='Delete'><input id='id' type='hidden' name='id' value=".$row['ID']."></form></td>"; 
    echo "</tr>"; 
    } 
    echo "</table>"; 
} 
else 
    echo "Save Failed!"; 
//close connection to database 
mysqli_close($con); 
?>

刪除頁：

<?php 
//create connection to Database 
$hostname = "localhost"; 
$database = "test"; 
$username = "root"; 
$password = ""; 
$con = mysqli_connect($hostname, $username, $password) or trigger_error(mysql_error(),E_USER_ERROR); 
mysqli_select_db($con,$database); 
//retrieve posted variables 
$id= mysql_real_escape_string($_POST['id']); 

$sql="DELETE FROM info WHERE ID = $id"; 
$query = mysql_query($con,$sql); 
if(mysql_affected_rows()>0){ 
$sql2="SELECT * FROM info ORDER BY ID ASC"; 
    $result = mysqli_query($con,$sql2); 
    echo "<table border='1'> 
    <tr> 
    <th>Name</th> 
    <th>Comment</th> 
    <th>Remove Entry</th> 
    </tr>"; 

    while($row = mysqli_fetch_array($result)) 
    { 
    echo "<tr>"; 
    echo "<td>" . $row['NAME'] . "</td>"; 
    echo "<td>" . $row['COMMENTS'] . "</td>"; 
    echo "<td> <form id='myForm2 method='post' action='delete.php'><input name='del' id='del' type='button' value='Delete'><input id='id' type='hidden' name='id' value=".$row['ID']."></form></td>"; 
    echo "</tr>"; 
    } 
    echo "</table>"; 
}else{ 
echo "Error deleting Data"; 
} 
mysqli_close($con); 
?>

腳本：

$(document).ready(function() 
{ 
    $("#sub").click(function() 
    { 
     //Check to see that no fields are left blank if one is all fields are cleared and save is failed from insert.php. 
     if ($("#name").val()=="") 
     { 
      clearInput(); 
      return false; 
     } 
     else if($("#comment").val()=="") 
     { 
      clearInput(); 
      return false; 
     } 
     //save the fields into an array and post to the result span then clear inputs 
     $.post($("#myForm").attr("action"), 
     $("#myForm :input").serializeArray(), 
     function(info){ $("#result").html(info);}); 
     clearInput(); 
    }); 

    $('#del').click(function() 
    { 
     $.post("delete.php",$("#myForm2 :input").serialize(),function(info){ $("#result").html(info);}); 
    }); 
}); 

$("#myForm").submit(function() { 
    return false; 
});

來源

2013-12-09 FearlessRooster

如果你不得不張貼一段代碼，那麼你不會努力去調試它。 –

*旁註：*您同時使用'mysql_ *'和'mysqli_ *'庫！爲什麼？ – Raptor

螢火蟲怎麼說？檢查請求的響應。 – Zarathuztra

由於輸入是在循環內創建的，因此您有多個具有相同ID del的元素 - 當您使用id選擇器時，它將僅返回具有給定id的第一個元素，因此您的點擊處理程序將僅註冊到第一個元素元件。您可以使用class屬性來分組類似的元素而不是id。

<input name='del' id='del' type='button' value='Delete'>變化到

<input name='del' class='del' type='button' value='Delete'>

然後使用類選擇來註冊處理程序。

$('#del').click(function() { 
    $.post("delete.php", $("#myForm2 :input").serialize(), function (info) { 
     $("#result").html(info); 
    }); 
});

來源

2013-12-09 03:00:49

謝謝你幫助。 – FearlessRooster

爲什麼我的動態刪除不起作用？

回答

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