怎么在表格中添加表单，并提交MYSQL数据库

如何在表格中添加表单，并提交MYSQL数据库
本人是个新手，现在想做个指标提交系统，MYSQL数据库中已经有销售整体架构，并已经在PHP中按照登陆账号自动生成对应的销售下属人员，但是架构后面跟着销售的指标，想做个输入框，并提交到数据库中，请问怎么做：

架构                                      盈利额
-L4:上海结婚 黄主管             输入框
-L3:上海结婚2组 沈主管       输入框
-L2:上海结婚2-1组 马主管    输入框
L1: 孙销售                            输入框
L1: 杜销售                            输入框
L1: 李销售                            输入框

<br /><?php <br />$q=$_POST["employeenumber"];<br /><br />$con = mysql_connect('localhost', 'root', '');<br />if (!$con)<br /> {<br /> die('Could not connect: ' . mysql_error());<br /> }<br /><br />mysql_select_db("org", $con);<br /><br />echo "<table border='1' cellpadding='10'><br /><tr><br /><th>架构</th><br /><th>盈利额</th><br /></tr>";<br /><br />$sql="SELECT 架构,盈利额 FROM `org` WHERE 主管工号 = '".$q."'";<br /><br />$result = mysql_query($sql);<br /><br />while($row = mysql_fetch_array($result))<br /> {<br /> echo "<tr>";<br /> echo "<td>" . $row['架构'] . "</td>";<br /> echo "</tr>";<br /> }<br />echo "</table>";<br /><br />mysql_close($con);<br />?><br>

------解决思路----------------------
我在你基礎上修改了一下，思路就是這樣，用表記錄的id對應提交。
index.php

<br /><?php <br />$q=$_POST["employeenumber"];<br /> <br />$con = mysql_connect('localhost', 'root', '');<br />if (!$con)<br /> {<br /> die('Could not connect: ' . mysql_error());<br /> }<br /> <br />mysql_select_db("org", $con);<br /><br />echo '<form name="form1" method="post" action="add.php">';<br />echo "<table border='1' cellpadding='10'><br /><tr><br /><th>架构</th><br /><th>盈利额</th><br /></tr>";<br /> <br />$sql="SELECT id, 架构,盈利额 FROM `org` WHERE 主管工号 = '".$q."'";<br /> <br />$result = mysql_query($sql);<br /> <br />while($row = mysql_fetch_array($result))<br /> {<br /> echo "<tr>";<br /> echo "<td>" . $row['架构'] . "</td>";<br /> echo '<td><input type="text" name="yl'.$row['id'].'"></td>';<br /> echo "</tr>";<br /> }<br />echo "</table>";<br />echo '<input type="hidden" name="employeenumber" value="'.$q.'">';<br />echo '</form>'; <br />mysql_close($con);<br />?><br />

add.php

<br /><?php<br />$con = mysql_connect('localhost', 'root', '');<br />if (!$con)<br /> {<br /> die('Could not connect: ' . mysql_error());<br /> }<br /> <br />mysql_select_db("org", $con);<br /><br />$employeenumber = $_POST["employeenumber"];<br /><br />$sql="SELECT id, 架构,盈利额 FROM `org` WHERE 主管工号 = '".$q."'";<br /> <br />$result = mysql_query($sql);<br /> <br />while($row = mysql_fetch_array($result))<br />{<br />	if($_POST['yl'.$row['id']]){<br />		$sqlstr = "update `org` set 盈利额='".$_POST['yl'.$row['id']]."' where id='".$row['id']."'"; // 更新入db<br />		mysql_query($sqlstr) or die(mysql_error());<br />	}<br />}<br /><br />mysql_close($con);<br /><br />header('location:index.php?q='.$employeenumber); // 跳轉回去<br />?><br />

------解决思路----------------------
24行的
盈利额='"$_POST['yl'.$row['id'].]"' where
改为
盈利额='".$_POST['yl'.$row['id']]."' where


------解决思路----------------------

本帖最后由 xuzuning 于 2014-12-10 18:33:58 编辑

$sqlstr = "update `org` set 盈利额='"$_POST['yl'.$row['id'].]"' where id='".$row['id']."'";
应为

$sqlstr = "update `org` set 盈利额='" . $_POST['yl'.$row['id']] . "' where id='".$row['id']."'";

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