1

目标

我正在尝试创建一个页面以从我的数据库中删除记录。该页面将由一个在页面加载时自行填充的页面组成。从 中选择一个值并单击提交按钮后，将调用一个 php 页面并显示结果php 将被加载到下面的表中。然后我可以单击位于回显值旁边的删除按钮以从数据库中删除该值

我的表格：

<!DOCTYPE HTML>
<html>
<head>
<!--Loads JQuery script-->
<script src="//code.jquery.com/jquery-1.9.1.js"></script>
<!--Gets list of item categories on page load-->
<script type="text/javascript">
$(document).ready(function(){
    $("#viewsubcat").load("getcategory.php");
});
</script>
<script type="text/javascript">
$("#viewsubcatsubmit").click(function(){
    var cat=$('#viewsubcat').val();
    $('#deletetable').load('delsubcategory.php?cat='+cat);
});
</script>
</head>
<body>

<form style="width:500px" id="viewsubcategory" name="viewsubcategory" method="post" action="<? php echo $_SERVER['PHP_SELF'] ?>" >
    <div class="inputfield">
        <label for="viewsubcat">Select Category:</label>
        <select style="margin-left:37px" id="viewsubcat" name="viewsubcat"></select>
    </div><br />

    <div class="inputfield">
        <input style="margin-left:250px" type="button" id="viewsubcatsubmit" name="viewsubcatsubmit" value="Search" /></div>
    </div><br />
</form>

<table id="deletetable">
</table>

</body>
</html>

PHP页面：

<?php
    include("cxn.inc");
    $id=$_SESSION['BizID'];
    $cat=$_GET['cat'];
    $viewsubcat=$cxn->prepare("SELECT * FROM `itemcat`,`itemsubcat` 
    WHERE `itemcat`.`CatID`=:cat AND `itemsubcat`.`ItemCat`=:cat AND `itemsubcat`.`BusinessID`=:id");
    $viewsubcat->bindValue(":cat",$cat);
    $viewsubcat->bindValue(":id",$id);
    $viewsubcat->execute();
    //echo"<table border='1'>";
    echo"<tr>";
            echo"<td>";
                echo"Categories";
            echo"</td>";
            echo"<td>";
                echo"SubCategories";
            echo"</td>";
            echo"<td>";
                echo"Action";
            echo"</td>";
        echo"</tr>";
    while($getsubcat=$viewsubcat->fetch(PDO::FETCH_ASSOC))
    {
        $cat=$getsubcat['ItemCat'];
        $subcat=$getsubcat['ItemSubCat'];
        $subcatid=$getsubcat['SubCatID'];
        echo"<tr>";
            echo"<td>";
                echo"$cat";
            echo"</td>";
            echo"<td>";
                echo"$subcat";
            echo"</td>";
            echo"<td>";
                echo"<form id='delsubcategory' name='delsubcategory' method='POST' action='delsubcategory.php'>";
                    echo"<input type='hidden' id='delsubcatid' name='delsubcatid' value='$subcatid' />";
                    echo"<input type='submit' id='delsubcatsubmit' name='delsubcatsubmit' value='Delete' />";
                echo"</form>";
            echo"</td>";
        echo"</tr>";
    }
    //echo"</table>";
?>

问题：

按下按钮后 HTML 表未加载。当我将表单的操作更改为“delsubcategory.php”并且按钮类型 =“submit”时，Php 页面正在工作，因此问题出在 html 上。

希望对此事有任何见解

4

1 回答 1

1

您忘记将按钮单击事件处理程序包装到$(function(){ });

<script type="text/javascript">
$(function(){
$("#viewsubcatsubmit").click(function(){
    var cat=$('#viewsubcat').val();
    $('#deletetable').load('delsubcategory.php?cat='+cat);
});
});
</script>
于 2013-08-26T04:58:04.003 回答