php 如何从html表单操作调用php函数？
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How to call php function from html form action?
提问by trendzcreator
I want to call php function in form action and i want to pass id as a argument. What I am doing is, in html form database column values will be displayed in text boxes, If I edit those values and click 'update' button values in database should be updated and 'Record updated successfully'message should be displayed in same page. I tried below code but not working. Let me know the solution. Thanks in advance.
我想在表单操作中调用 php 函数，我想将 id 作为参数传递。我正在做的是，在 html 表单中，数据库列值将显示在文本框中，如果我编辑这些值并单击“更新”按钮，数据库中的值应该被更新，并且“记录更新成功”消息应该显示在同一页面中。我试过下面的代码但没有用。让我知道解决方案。提前致谢。
<html>
<head>
<link rel="stylesheet" type="text/css" href="cms_style.css">
</head>
<?php
$ResumeID = $_GET['id'];
$con = mysql_connect("localhost", "root", "");
mysql_select_db("engg",$con);
$sql="SELECT * from data WHERE ResumeID=$ResumeID";
$result = mysql_query($sql);
$Row=mysql_fetch_row($result);
function updateRecord()
{
//If(!isset($_GET['id']))
//{
$NameoftheCandidate=$_POST[NameoftheCandidate];
$TelephoneNo=$_POST[TelephoneNo];
$Email=$_POST[Email];
$sql="UPDATE data SET NameoftheCandidate='$_POST[NameoftheCandidate]', TelephoneNo='$_POST[TelephoneNo]', Email='$_POST[Email]' WHERE ResumeID=$ResumeID ";
if(mysql_query($sql))
echo "<p>Record updated Successfully</p>";
else
echo "<p>Record update failed</p>";
while ($Row=mysql_fetch_array($result)) {
echo ("<td>$Row[ResumeID]</td>");
echo ("<td>$Row[NameoftheCandidate]</td>");
echo ("<td>$Row[TelephoneNo]</td>");
echo ("<td>$Row[Email]</td>");
} // end of while
} // end of update function
?>
<body>
<h2 align="center">Update the Record</h2>
<form align="center" action="updateRecord()" method="post">
<table align="center">
<input type="hidden" name="resumeid" value="<? echo "$Row[1]"?>">
<? echo "<tr> <td> Resume ID </td> <td>$Row[1]</td> </tr>" ?>
<div align="center">
<tr>
<td> Name of the Candidate</td>
<td><input type="text" name="NameoftheCandidate"
size="25" value="<? echo "$Row[0]"? >"></td>
</tr>
<tr>
<td>TelephoneNo</td>
<td><input type="text" name="TelephoneNo" size="25" value="<? echo "$Row[1]"?>"></td>
</tr>
<tr>
<td>Email</td>
<td><input type="text" name="Email" size="25" value="<? echo "$Row[3]"?>">
</td>
</tr>
<tr>
<td></td>
<td align="center"><input type="submit" name="submitvalue" value="UPDATE" ></td>
</tr>
</div>
</table>
</form>
</body>
</html>
回答by Dino
try this way
试试这个方法
HTML
HTML
<form align="center" action="yourpage.php?func_name=updateRecord" method="post">
PHP
PHP
$form_action_func = $_POST['func_name'];
if (function_exists($form_action_func)) {
updateRecord();
}
回答by Butani Vijay
write form action="" and then write your php code as below note : use form method as get
写form action =“”然后写你的php代码如下注意：使用form方法作为get
<?php
if(isset($_GET['id']))
{
call your function here
}
?>
in function access all values using $_GET['fieldname']
在函数中使用 $_GET['fieldname'] 访问所有值
回答by Sanket Shembekar
simple way make your "Submit " and "Update" action performed on same page then
简单的方法让你的“提交”和“更新”操作在同一页面上执行然后
if(isset($_POST['update']))
{
//perform update task
update($var1,var2,$etc); // pass variables to function
header('Location: http://www.example.com/');// link to your form
}
else if(isset($_POST['submit']))
{
//perform update task
submit($var1,$var2,$etc);// pass variables to function
header('Location: http://www.example.com/'); // link to next page after submit successfully
}
else
{
// display form
}
