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我有一個表格，其中包含一個可以通過單擊添加按鈕動態添加行的部分。和往常一樣，爲什麼我要來這裏將添加的行添加到數據庫中時遇到問題。我的意思是，動態添加行的值不會被保存到數據庫中，只會添加第一行的值。如何將動態添加的行的值存入數據庫？
這是腳本：
編輯：
<script language="Javascript">
var i=1;
function addRowToTable()
{
i++;
m.r.value = i;
var tbl = document.getElementById('table');
var lastRow = tbl.rows.length;
var iteration = lastRow;
var row = tbl.insertRow(lastRow);
var cellLeft = row.insertCell(0);
var textNode = document.createTextNode(iteration);
cellLeft.appendChild(textNode);
var cellRightSel1 = row.insertCell(1);
var sel = document.createElement('select');
sel.name = 'name' + iteration;
sel.setAttribute("onchange", "choosec(this);");
var item = new Option("","");
sel.options[sel.length] = item;
<?
while($data = mysql_fetch_array($result)){
?>
var item = new Option("<?=$data["Name"];?>","<?=$data["ID"];?>");
sel.options[sel.length] = item;
<? } ?>
cellRightSel1.appendChild(sel);
var cellRightSel2 = row.insertCell(2);
var sel = document.createElement('select');
sel.name = 'class' + iteration;
sel.setAttribute("onchange", "choosepoint(this);");
var item = new Option ("","");
sel.options[sel.length] = item;
<?
while($data = mysql_fetch_array($result_sub)){
?>
var item = new Option("<?=$data["Class"];?>","<?=$data["ID"];?>");
sel.options[sel.length] = item;
<? } ?>
cellRightSel2.appendChild(sel);
var cellRight = row.insertCell(3);
var div = document.createElement('div');
div.id = 'point' + iteration;
cellRight.appendChild(div);
}
</script>
<form id="m" name="m" method="POST" action="submit.php">
<input type="hidden" name="menu_id" value="<?=$menu_id;?>">
<input type="hidden" name="r" id="r">
<table align="center" class="table">
<tr>
<tr>
<td class="tablesubtitle">Adjustment</td>
<td class="tablesubtitle"><input type="text" name="adjustment" size="2"></td>
</td>
</tr>
<tr>
<td class="tablesubtitle">Component</td>
<td class="tablesubtitle">
<table id="table">
<tr>
<td>No</td>
<td>Name</td>
<td>Class</td>
<td>Point</td>
<td>Action</td>
</tr>
<tr>
<td>1</td>
<td><select name="name">
<option value="#"> </option>
<?php
opendb();
$query = "SELECT * FROM student";
$result = mysql_query($query);
while($data = mysql_fetch_array($result)){
$ID = $data['ID'];
$Name = $data['Name'];
echo "<option value='$ID'>$Name</option>";
}
?>
</select>
</td>
<td><select name='class' onchange="choosepoint(this);">
<option value="#"> </option>
<?php
opendb();
$query = "SELECT * FROM student2";
$result = mysql_query($query);
while($data = mysql_fetch_array($result)){
$ID1 = $data['ID'];
$class = $data['Class'];
echo "<option value='$ID'>$class</option>";
}
?>
</select>
</td>
<td>
<div id="point" name="point"></div>
</td>
<td><input type="button" value=" + " onClick="addRowToTable();"> | <input type="button" value=" - " onClick="removeRowFromTable();"></td>
</tr>
</table>
</td>
<tr>
<td></td>
<td align="center"><input type="submit" value="Submit"></td>
</tr>
</tr>
</table>
</form>
Submit.php
<?php
include ("c/config.php");
include ("c/db.php");
opendb();
$query1 = "select * from student where ID='$name'";
$result1 = querydb($query1);
$data = mysql_fetch_array($result1);
$adjustment = $data['adjustment'];
$num = $_POST['r'];
//echo $num;
for($i=0; $i<=$num; $i++){
if(isset($_REQUEST['name$i'])){
$name = $_REQUEST['name$i'];
}
if(isset($_REQUEST['class$i'])){
$class = $_REQUEST['class$i'];
}
if(isset($_REQUEST['point$i'])){
$point = $_REQUEST['point$i'];
}
$query2 = "INSERT INTO student3
VALUES('$adjustment','$name','$class','$point')";
$result2 = querydb($query2);
}
closedb();
?>
。在你的代碼中沒有插入或更新語句。另外這裏的代碼太多了。減少代碼量，只發布給你帶來麻煩的部分。 – Dan 2012-02-07 08:18:10
呃，我忘記了，我很抱歉。 – candies 2012-02-07 08:33:57