我想分别获取 2 个值,以便通过 AJAX 从 PHP 自动填充第一个文件中的输入,但是当我使用 console.log(data) 时,我获取了所有数据并且我想单独检索它为了把它放在不同的标签中
<input id="regist" type="text" name="regist" onblur="passdata();">
<input id="atyp" type="text" name="atyp" >
<input id="mtow" type="text" name="mtow" >
<script>
$(document).ready(function(){})
function passdata() {
var regist = $('#regist').val();
$.ajax({
type: 'post',
url: 'checkdb.php',
data: 'regist='+regist,
success: function(data){
var atyp = $('#myvalue1').val();
var mtow = $('#myvalue1').val();
$('#atyp').text(atyp);
$('#mtow').text(mtow);
alert (aty+mtow);
console.log(data);
}
})
}
</script>
和 PHP 文件....当然有数据库连接
$regist = $_POST['regist'];
$conn = mysqli_connect($host,$user,$pwd,$db);
$sql= ("SELECT * FROM aircrafts where regist='$regist'");
$datas = mysqli_query($conn,$sql);
foreach ($datas as $row) {};
if(mysqli_num_rows($datas) == 0) {
echo 'non ce niente';
} else { ?>
<span id="myvalue1"><?php echo $atyp = $row['atyp'];?></span>
<span id="myvalue2"><?php echo $atyp = $row['mtow'];?></span>
<?php };
?>
最佳答案
将您的 Ajax 请求更改为:
$.ajax({
type: 'post',
url: 'checkdb.php',
data: 'regist='+regist,
dataType: 'json', // NEW LINE
success: function(data) {
//var atyp = $('#myvalue1').val();
//var mtow = $('#myvalue1').val();
//$('#atyp').text(atyp);
//$('#mtow').text(mtow);
var atyp = data.atyp;
var mtow = data.mtow;
alert ('aty: ' + aty , 'mtow: ' + mtow);
console.log(data);
}
})
并在您的 PHP 中:更改这些行
else{?>
<span id="myvalue1"><?php echo $atyp = $row['atyp'];?></span>
<span id="myvalue2"><?php echo $atyp = $row['mtow'];?></span>
<?php };
正好是这条线
else {
echo json_encode( $row );
}
编辑
关于您的“DOM 事件”处理(“document onready”和“input#regist onblur”)我建议这样做。
HTML:
<!-- <input id="regist" type="text" name="regist" onblur="passdata();"> -->
<input id="regist" type="text" name="regist">
<input id="atyp" type="text" name="atyp">
<input id="mtow" type="text" name="mtow">
JavaScript:
function passdata(event) {
$.ajax({
type: 'post',
url: 'checkdb.php',
data: 'regist=' + $(event.target).val(),
dataType: 'json',
success: function(data){
console.log(data);
}
})
}
$(document).ready(function(){
// on blur event
$('#regist').on('blur', passdata)
// on input event | maybe this would be also an interesting option as it fires immidiately on input
// $('#regist').on('input', passdata)
})
此 Bootstrap 将逻辑与标记分开,这通常是一种很好的做法。
关于javascript - 如何使用 AJAX 从 php 获取值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46402064/