ajax和php输入多个表单输入到数据库我有一个PHP生成的表单，其中包含多个输入字段，其数量由用户选择确定。我想使用ajax函数将所有数据输入数据库。问题是我是ajax的新手，并不确定如何去做。下面的ajax javascript函数是我想要实现的一个示例，我知道它不正确。有人能指出我正确的方向。我环顾四周，从我看到的情况来看，Json可能是一个解决方案，但我对它一无所知并且阅读它我仍然没有得到它。样本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;
}表格样本：<?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++;
}
?>
2 回答
千巷猫影
TA贡献1829条经验 获得超7个赞
这是一个简单的AJAX演示：
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的
// 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'));
}
?>
红糖糍粑
TA贡献1815条经验 获得超6个赞
var postdata={};postdata['num']=x;
while (i <= x){
postdata['fname'+i]= name[i];
postdata['lname'+i]= lname[i];
postdata['email'+i]= email[i];
i++;}
$.ajax({url : 'process.php',
type:"POST",
data:postdata,
success : function(data){
window.setTimeout(function()
{
$('#SuccessDiv').html('Info Added!');
$('#data').css("display","block");
$('#data').html(data);
}, 2000);
}});PHP
$num=$_POST['num'];for($i=1;i<=$num;i++){
echo $_POST['fname'.$i];
echo $_POST['lname'.$i];
echo $_POST['email'.$i];}- 2 回答
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