[英]How post more than one variable in javascript?
There is 2 input type="text". 有2个输入type =“ text”。 First, user input 1st input text area with id="ncr_no".
首先,用户输入的第一个输入文本区域的ID =“ ncr_no”。 Then, cursor is in 2nd input type "text" with id="itm_cd".
然后,光标位于id =“ itm_cd”的第二个输入类型“ text”中。 Now, I want to make, how the two input by user, when cursor is in 2nd input type, posted to other php (get_ncrnoitmcd.php) by javascript?
现在,我要制作的是当光标处于第二输入类型时,用户的两个输入如何通过javascript发布到其他php(get_ncrnoitmcd.php)? That's the code.
那就是代码。
<script type="text/javascript">
$(document).ready(function() {
$("#itm_cd").keyup(function (e) {
$(this).val($(this).val().replace(/\s/g, ''));
var itm_cd = $(this).val();
if(itm_cd.length < 1){$("#user-result3").html('');return;}
if(itm_cd.length >= 1 ){
$("#user-result3").html('<img src="image/ajax-loader.gif" />');
$.post('get_ncrnoitmcd.php', {'itm_cd':itm_cd}, function(data) {
$("#user-result3").html(data);
});
}
});
});
</script>
Thank a lot. 非常感谢。
This is the way you can send the 2 values to server on 2nd element keyup after validation. 这样,您便可以在验证后将2个值发送到第二个元素keyup上的服务器。 Whats the problem that you are facing?
您面临的问题是什么? I also added ncr_no in the post request.
我还在发布请求中添加了ncr_no。
<script type="text/javascript">
$(document).ready(function() {
$("#itm_cd").keyup(function (e) {
$(this).val($(this).val().replace(/\s/g, ''));
var itm_cd = $(this).val();
if(itm_cd.length < 1){
$("#user-result3").html('');
return;
}else if(itm_cd.length >= 1 ){
$("#user-result3").html('<img src="image/ajax-loader.gif" />');
$.post(
'get_ncrnoitmcd.php'
,{'itm_cd':itm_cd,'ncr_no':$('#ncr_no').val()}
,function(data) {
$("#user-result3").html(data);
}
);
}
});
});
</script>
I want to get available or not available, Mr. @joyBlanks 我想上班还是不上班,@ joyBlanks先生
<?php
//connection.php
if(isset($_POST["itm_cd"],$_POST["ncr_no"]))
{
if(!isset($_SERVER['HTTP_X_REQUESTED_WITH']) AND strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) != 'xmlhttprequest') {
die();
}
$connecDB = mysqli_connect($db_host, $db_username, $db_password,$db_name)or die('could not connect to database');
$itm_cd = strtolower(trim($_POST["itm_cd"]));
$itm_cd = filter_var($itm_cd, FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_LOW|FILTER_FLAG_STRIP_HIGH);
$results = mysqli_query($connecDB,"SELECT itm_cd,ncr_no FROM sqc_ncr WHERE itm_cd ='$itm_cd' AND ncr_no ='$ncr_no'");
$itm_cd_exist = mysqli_num_rows($results);
if($itm_cd_exist) {
die('<!--img src="image/available.png" /--> <i>Available in database</i>');
}else{
die('<!--img src="image/not-available.png" /--> <i>Not Available in database</i>');
}
mysqli_close($connecDB);
}
?>
我没有使用过,也没有使用过&&,也没有使用过http_x_requested,但是在未显示的html中可用或不可用。
<td><input type="text" class="input" name="itm_cd" id="itm_cd" onBlur="updateItemName()" required /> <span id="user-result3"></span></td>
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