ajax和php输入多个表单输入到数据库 [英] ajax and php to enter multiple forms input to database
问题描述
我有一个 php 生成的表单,其中包含多个输入字段,其数量由用户选择决定.我想使用 ajax 函数将所有数据输入到数据库中.问题是我是 ajax 的新手,不确定如何去做.下面的 ajax javascript 函数是我想要实现的示例,我知道它不正确.有人可以指出我正确的方向.我环顾四周,从我所看到的来看,Json 可能是一个解决方案,但我对此一无所知,阅读它我仍然不明白.
I have a php generated form with multiple input fields the number of which is determined by user select. I would like to use an ajax function to enter all the data to database. The problem is that I am new to ajax and not sure ho to go about it. The ajax javascript function below is a sample of what I would like to achieve and I know it is not correct. Can someone point me in the right direction. I have looked around and from what I see, Json may be a solution but I know nothing about it and reading up on it I still don't get it.
示例ajax:
function MyFunction(){
var i = 1;
var x = $('#num_to_enter').val();
while (i <= x){
var name = $('#fname[i]').val();
var lname = $('#lname[i]').val();
var email = $('#Email[i]').val();
i++;
}
$('#SuccessDiv').html('Entering Info.<img src="images/processing.gif" />');
$.ajax({url : 'process.php',
type:"POST",
while (i <= x){
data: "fname[i]=" + name[i] + "&lname[i]=" + lname[i] + "&email[i]=" + email[i],
i++;
}
success : function(data){
window.setTimeout(function()
{
$('#SuccessDiv').html('Info Added!');
$('#data').css("display","block");
$('#data').html(data);
}, 2000);
}
});
return false;
}
表格示例:
<?php
echo "<form method='post'>";
$i=1;
while($i <= $num_to_enter){
$form_output .= "First Name:
<input id='fname' type='text' name='fname[$i]'><br />
Last Name:
<input id='lname' type='text' name='lname[$i]'><br />
Email:
<input id='Email' type='text' name='Email[$i]'><br />
$i++;
}
echo"<input type='button' value='SUBMIT' onClick='MyFunction()'></form>";
?>
Then DB MySQL Sample
<?php
while ($i <= $x){
$x = $_POST['num_to_enter'];
$fname = $_POST['fname[$i]'];
$fname = $_POST['fname[$i]'];
$fname = $_POST['email[$i]'];
$sql = "INSERT INTO `mytable`
(`firstname`, `lastname`, `email`) VALUES ('$fname[$i]', '$lname[$i]', '$email[$i]');";
$i++;
}
?>
推荐答案
这里是一个简单的AJAX演示:
Here is a simple demo of AJAX:
HTML
<form method="POST" action="process.php" id="my_form">
<input type="text" name="firstname[]">
<input type="text" name="firstname[]">
<input type="text" name="firstname[]">
<input type="text" name="firstname[custom1]">
<input type="text" name="firstname[custom2]">
<br><br>
<input type="submit" value="Submit">
</form>
jQuery
// listen for user to SUBMIT the form
$(document).on('submit', '#my_form', function(e){
// do not allow native browser submit process to proceed
e.preventDefault();
// AJAX yay!
$.ajax({
url: $(this).attr('action') // <- find process.php from action attribute
,async: true // <- don't hold things up
,cache: false // <- don't let cache issues haunt you
,type: $(this).attr('method') // <- find POST from method attribute
,data: $(this).serialize() // <- create the object to be POSTed to process.php
,dataType: 'json' // <- we expect JSON from the PHP file
,success: function(data){
// Server responded with a 200 code
// data is a JSON object so treat it as such
// un-comment below for debuggin goodness
// console.log(data);
if(data.success == 'yes'){
alert('yay!');
}
else{
alert('insert failed!');
}
}
,error: function(){
// There was an error such as the server returning a 404 or 500
// or maybe the URL is not reachable
}
,complete: function(){
// Always perform this action after success() and error()
// have been called
}
});
});
PHP process.php
<?php
/**************************************************/
/* Uncommenting in here will break the AJAX call */
/* Don't use AJAX and just submit the form normally to see this in action */
// see all your POST data
// echo '<pre>'.print_r($_POST, true).'</pre>';
// see the first names only
// echo $_POST['firstname'][0];
// echo $_POST['firstname'][1];
// echo $_POST['firstname'][2];
// echo $_POST['firstname']['custom1'];
// echo $_POST['firstname']['custom2'];
/**************************************************/
// some logic for sql insert, you can do this part
if($sql_logic == 'success'){
// give JSON back to AJAX call
echo json_encode(array('success'=>'yes'));
}
else{
// give JSON back to AJAX call
echo json_encode(array('success'=>'no'));
}
?>
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