javascript 类型错误：e 未定义
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Type error : e is undefined
提问by Ajit
I am getting a type error: e undefined on the javascript code. I am trying to populate a dropdown list using jquery with data sent from a mysql server.
我收到一个类型错误：e 在 javascript 代码上未定义。我正在尝试使用 jquery 和从 mysql 服务器发送的数据填充下拉列表。
here is the javascript code
这是javascript代码
$(document).ready(function(){// This script uses jquery and ajax it is used to set the values in
$("#day").change(function(){// the time field whenever a day is selected.
var day=$("#day").val();
var doctor=$("#doctor").val();
$.ajax({
type:"post",
url:"time.php",
data:"day="+day+"&doctor="+doctor,
dataType : 'json',
success:function(data){
var option = '';
$.each(data.d, function(index, value) {
console.log(data.d);
option += '<option>' + value.arr + '</option>';
});
$('#timing').html(option);
}
});
});
});
Here's the php script which fetches data from a mysql database
这是从 mysql 数据库中获取数据的 php 脚本
$doctor = $_POST['doctor'];
$day = $_POST['day'];
$query="SELECT * FROM schedule WHERE doctor='$doctor' AND day='$day'";
$arr = array();
$result = mysqli_query($con, $query);
$i = 0;
//Initialize the variable which passes over the array key values
$row = mysqli_fetch_assoc($result); //Fetches an associative array of the row
$index = array_keys($row); // Fetches an array of keys for the row.
while($row[$index[$i]] != NULL)
{
if($row[$index[$i]] == 1) {
//$res = $index[$i];
//echo json_encode($res);
array_push($arr, $index[$res]);
}
$i++;
}
回答by Rohan Kumar
You are not using each()in a right way replace the line
您没有以正确的方式使用each()替换该行
$.each(data.d, function(index, value) {
by
经过
if(data.d)
{
$(data.d).each(function(index, value) {
// your code
Alsoin phpuse json_encode()to make your dataas json
同样在php 中使用json_encode()将您的数据设为json
} // end of while
echo json_encode($arr);
Updated code,try in your php script,
更新代码，在你的php 脚本中尝试，
$i = 0;
//Initialize the variable which passes over the array key values
while($row = mysqli_fetch_assoc($result))
{
$arr['d'][$i]=$row['doctor'];
// you can add more fields in array like above
$i++;
}
echo json_encode($arr);
return;
Andin your Javascriptit will work after reading DOC jquery.each()
而在你的JavaScript它会读取DOC后工作jquery.each（）
$.each(data.d, function(index, value) {
option += '<option>' + value+ '</option>';
});
回答by Jacob Clark
It appears you're not parsing your JSON, try the following, note the line
看来您没有解析您的 JSON，请尝试以下操作，注意该行
data = JSON.parse(data);
data = JSON.parse(data);
You must use JSON.parse on your data object otherwise JavaScript will treat it as a string.
您必须在数据对象上使用 JSON.parse，否则 JavaScript 会将其视为字符串。
$(document).ready(function(){// This script uses jquery and ajax it is used to set the values in
$("#day").change(function(){// the time field whenever a day is selected.
var day=$("#day").val();
var doctor=$("#doctor").val();
$.ajax({
type:"post",
url:"time.php",
data:"day="+day+"&doctor="+doctor,
dataType : 'json',
success:function(data){
var option = '';
data = JSON.parse(data);
$.each(data.d, function(index, value) {
console.log(data.d);
option += '<option>' + value.arr + '</option>';
});
$('#timing').html(option);
}
});
