how can i alert in javascript the value echo from php
NOTE I TRIED THE ALERT(DATA) BUT IT din't give any validation when success because of the lopping thing in php
my concern is when i try making validation it just go straight to else thing even thought which is Import Successfully even if the data or the echo in php is "Error1" , "Error2" or "Error3" please help..
here is my ajax
<script>
$(document).ready(function(){
$('#upload_csv').on("submit", function(e){
e.preventDefault(); //form will not submitted
$.ajax({
url:"import.php",
method:"POST",
data:new FormData(this),
contentType:false, // The content type used when sending data to the server.
cache:false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data){
if(data == 'Error1')
{
alert("Error1");
}
else if(data == 'Error2')
{
alert("Error2");
}
else if(data == 'Error3')
{
alert('Error3');
}
else
{
alert('Import Successfull');
$('#employee_table').html(data);
}
}
})
});
});
</script>
and here is the php code
<?php
if(!empty($_FILES["employee_file"]["name"]))
{
$connect = mysqli_connect("localhost", "root", "", "db");
$output = '';
$allowed_ext = array("csv");
$tmp = explode(".", $_FILES["employee_file"]["name"]);
$extension = end($tmp);
if(in_array($extension, $allowed_ext))
{
$file_data = fopen($_FILES["employee_file"]["tmp_name"], 'r');
fgetcsv($file_data);
while($row = fgetcsv($file_data))
{
$try = count($row);
if($try === 6){
$id = mysqli_real_escape_string($connect, $row[0]);
$name = mysqli_real_escape_string($connect, $row[1]);
$address = mysqli_real_escape_string($connect, $row[2]);
$gender = mysqli_real_escape_string($connect, $row[3]);
$designation = mysqli_real_escape_string($connect, $row[4]);
$age = mysqli_real_escape_string($connect, $row[5]);
$query = "
INSERT INTO tbl_employee
(id,name, address, gender, designation, age)
VALUES ('$id','$name', '$address', '$gender', '$designation', '$age')
";
mysqli_query($connect, $query);
}else
{
echo 'Error3';
break;
}
}
echo $output;
}
else
{
echo "Error1";
}
}
else
{
echo "Error2";
}
?>
alter()
cannot output complex data. try to use console.log()
.
data will appear in the browser console instead of popping up directly. You can press F12 and then click on the Console to view
try the return value using json, on js
<script>
$(document).ready(function(){
$('#upload_csv').on("submit", function(e){
e.preventDefault(); //form will not submitted
$.ajax({
url:"import.php",
method:"POST",
dataType: 'JSON',
data:new FormData(this),
contentType:false, // The content type used when sending data to the server.
cache:false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data){
// look into detail json data
console.log(data);
if(data['status'] == false)
{
alert(data['error']);
} else
{
alert('Import Successfull');
$('#employee_table').html(data);
}
}
})
});
});
</script>
on php
<?php
if(!empty($_FILES["employee_file"]["name"]))
{
$connect = mysqli_connect("localhost", "root", "", "db");
$output = '';
$allowed_ext = array("csv");
$tmp = explode(".", $_FILES["employee_file"]["name"]);
$extension = end($tmp);
if(in_array($extension, $allowed_ext))
{
$file_data = fopen($_FILES["employee_file"]["tmp_name"], 'r');
fgetcsv($file_data);
while($row = fgetcsv($file_data))
{
$try = count($row);
if($try === 6){
$id = mysqli_real_escape_string($connect, $row[0]);
$name = mysqli_real_escape_string($connect, $row[1]);
$address = mysqli_real_escape_string($connect, $row[2]);
$gender = mysqli_real_escape_string($connect, $row[3]);
$designation = mysqli_real_escape_string($connect, $row[4]);
$age = mysqli_real_escape_string($connect, $row[5]);
$query = "INSERT INTO tbl_employee
(id,name, address, gender, designation, age)
VALUES ('$id','$name', '$address', '$gender', '$designation', '$age')";
mysqli_query($connect, $query);
}else{
$status = false;
$error = 'Error3';
$output = [];
}
}
$status = true;
$error = [];
$output = $output;
} else {
$status = false;
$error = "Error1";
$output = [];
}
} else {
$status = false;
$error = "Error2";
$output = [];
}
header('Content-Type: application/json');
echo json_encode(array('status'=>$status,'output'=>$output,'error'=>$error), JSON_PRETTY_PRINT);
?>
sorry for my bad english
使用console.log()测试您的数据,警报无法获取很多数据
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