[英]How to get value from select option without selected the option in Laravel
[英]How to get selected value of SELECT OPTION
我的代碼有問題,我有兩個選擇,第一個是“ drop_drop”,第二個是“ salidhan_ini”,每次更改“ drop_drop”時,它都會更改“ salidhan_ini”的選項,將使用php從數據庫保存“ salidhan_ini”的值,但我的代碼無法正常工作。
的index.php
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<script src="jquery-1.8.3.min.js"></script>
<script src="jquery-1.11.1.min.js"></script>
<script>
$(document).ready( function() {
$('#drop_drop').on('change', function (e) {
$.ajax({
type: 'POST',
url:"process.php",
data:{keyname:$('#drop_drop option:selected').val()},
success: function(simbag){
$("#salidhan_ini").html(simbag);
}
});
});
});
</script>
</head>
<body>
<div id="dropdown1">
<select name="taskOption" id="drop_drop">
<option value="1">What</option>
<option value="2">When</option>
<option value="3">Where</option>
</select>
</div>
<div id="dropdown2">
<select id="salidhan_ini">
</select>
</div>
</body>
</html>
process.php
<?php
$id_man = $_POSt['taskOption'];
$con=mysqli_connect("localhost","root","","sample_sample");
// Check connection
if (mysqli_connect_errno()){
die ("Failed to connect to MySQL: " . mysqli_connect_error());
}
$result = mysqli_query($con,"SELECT * FROM sample_hatak where id = $id_man");
if(!$result){
echo 'Error: '. mysql_error();
}else{
while($row = mysqli_fetch_array($result))
{
echo "<option value=". $row['FirstName'] .">". $row['LastName'] ."</option>";
}
}
mysqli_close($con);
?>
我想實現的是每次“ drop_drop”更改時,“ drop_drop”中所選選項的值將被粘貼到“ process.php”中,然后保存所有具有相應ID(即所選值)的數據選項“ drop_drop”,然后它將全部返回為“ salidhan_ini”的選項。 感謝您的幫助!
您需要填充第二選擇,如下所示:
$(document).ready( function() {
$('#drop_drop').on('change', function (e) {
$.ajax({
type: 'POST',
url:"process.php",
data:{keyname:$('#drop_drop option:selected').val()},
success: function(simbag){
$.each(simbag, function(val, text) {
$("#salidhan_ini").append( $('<option></option>').val(val).html(text) )
});
}
});
});
});
還僅包含一個版本的jQuery
感謝幫助! 我發現了我的問題,這個問題在process.php
<?php
$id_man = $_POSt['taskOption'];
$con=mysqli_connect("localhost","root","","sample_sample");
// Check connection
if (mysqli_connect_errno()){
die ("Failed to connect to MySQL: " . mysqli_connect_error());
}
$result = mysqli_query($con,"SELECT * FROM sample_hatak where id = $id_man");
if(!$result){
echo 'Error: '. mysql_error();
}else{
while($row = mysqli_fetch_array($result))
{
echo "<option value=". $row['FirstName'] .">". $row['LastName'] ."</option>";
}
}
mysqli_close($con);
?>
問題是“ $POSt
”,所以我將其更改為“ $_POST
”,
並且此echo "<option value=". $row['FirstName'] .">". $row['LastName']
echo "<option value=". $row['FirstName'] .">". $row['LastName']
echo "<option value=". $row['FirstName'] .">". $row['LastName']
FirstName
和LastName
不是我數據庫中屬性的一部分,因為它是firstname
和lastName
。
謝謝!
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