2013-08-26 35 views
1

目的加載到PHP的HTML表格通過jQuery的.load調用

我想創建一個網頁從我database.The頁面上刪除記錄將包括一個用於填充其本身在頁面load.Upon從中選擇一個值並點擊提交按鈕,一個php頁面將被調用,並且php的結果將被加載到一個下面的表格中。然後,我可以點擊刪除按鈕,該按鈕將被旁邊的值回顯以刪除從數據庫中值

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頁面:

<?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>"; 
?> 

問題:

HTML表沒有加載在按壓button.The PHP的頁面工作時我改變形式的行動 「delsubcategory.php」和按鈕類型=「提交」,所以問題在於html。

希望得到任何見解此事

+0

你得到一個錯誤? – Rab

+0

@RabNawaz Dipesh明白了。我忘了把我的.click換成$(function()); –

回答

1

你忘了你的包裹按鈕單擊事件處理程序到$(function(){ });

<script type="text/javascript"> 
$(function(){ 
$("#viewsubcatsubmit").click(function(){ 
    var cat=$('#viewsubcat').val(); 
    $('#deletetable').load('delsubcategory.php?cat='+cat); 
}); 
}); 
</script> 
+0

Arghh不能相信我錯過了。令人驚訝的是,最簡單的錯誤可能會導致數小時的頭髮拉動 –

+0

@Ken它發生在某個時候......但教你不要再重複一遍......希望問題解決了... –

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