I have a php.php file to receive data then pass the result back to AJAX, like this:
<?php
$p1 = $_POST["p1"];
$c1 = $_POST["c1"];
$resultx['r1'] = "XXX";
$resultx['r2'] = 0;
$resultx['r3'] = 100;
$resultx['r4'] = 1000;
echo json_encode($resultx);
And in my js file I have:
$.ajax({
url: 'php.php',
type: 'POST',
data: {
p1: 5,
c1: 100
},
cache: false,
success: function (resultx) {
var result = resultx;
console.log(result);
var my1 = document.getElementById('my1');
var my2 = document.getElementById('my2');
var my3 = document.getElementById('my3');
var my4 = document.getElementById('my4');
my1.textContent = "Id : "+result.r1;
my2.textContent = "Rank : "+result.r2;
my3.textContent = "Total players : "+result.r3;
my4.textContent = "Top credit : "+result.r4;
},
});
The my1 my2 my3 my4 fields all print with undefined even though the console log show the value of:
{"r1":"XXX","r2":0, "r3":100, "r4":1000}
When I change the statement
var result = resultx;
to
var result = JSON.parse(resultx)
the javascript console prints a error message of uncaught syntax error unexpected token.
Please help!
Related
I tried to call / search data by ID , from the server using ajax in cordova , but when I click to display the data " undefined" , what's wrong ???
This my html
<input type="text" id="result" value=""/>
<button>get</button>
<div id="result2"></div>
function get (){
var qrcode = document.getElementById ("result").value;
var dataString="qrcode="+qrcode;
$.ajax({
type: "GET",
url: "http://localhost/book/find.php",
crossDomain: true,
cache: false,
data: dataString,
success: function(result){
var result=$.parseJSON(result);
$.each(result, function(i, element){
var id_code=element.id_code;
var qrcode=element.qrcode;
var judul=element.judul;
var hasil2 =
"QR Code: " + qrcode + "<br>" +
"Judul: " + Judul;
document.getElementById("result2").innerHTML = hasil2;
});
}
});
}
This my script on server
include "db.php";
$qrcode= $_GET['qrcode'];
$array = array();
$result=mysql_query("select * from book WHERE qrcode LIKE '%{$qrcode}%'");
while ($row = mysql_fetch_array($result)) { //fetch the result from query into an array
{
$array[] =$row['qrcode'];
$array[] =$row['judul'];
$array[] =$row['jilid'];
}
echo json_encode($array);
}
try to change your parameter in calling ajax
var dataString = {qrcode : qrcode };
Change your php code as below
include "db.php";
$qrcode= $_GET['qrcode'];
$array = array();
$result=mysql_query("select * from book WHERE qrcode LIKE '%{$qrcode}%'");
while ($row = mysql_fetch_array($result)) { //fetch the result from query into an array
{
$array[] =['qrcode'=>$row['qrcode'],'judul'=>$row['judul'],'id_code'=>$row['jilid']];
}
echo json_encode($array);
}
RESOLVED the problem is. I try to replace with this script and the result was the same as I expected.
while ($row = mysql_fetch_object($result)){
$array[]=$row++;
}
echo json_encode($array);
I have a text box where I search database for a specific information.
The PHP code when I just type and click on search is the following:
try
{
$date_emp = $_POST['date_emp'];
$val = $_POST['data1'];
$gender = $_POST['gen'];
if($date_emp == "choose" && $gender == "specify")
{
$search = "SELECT * FROM employee
WHERE emp_name = :val OR position = :val
OR salary = :val OR date_employed = :val
OR gender = :val";
$searchStmt = $conn->prepare($search);
$searchStmt->bindValue(":val", $val);
$searchStmt->execute();
$res = $searchStmt->fetchAll();
echo json_encode($res);
}
catch(PDOException $ex)
{
echo $ex->getMessage();
}
And here the AJAX script for it:
$("#search").click(function()
{
var txt = $("#txtSearch").val();
var drop = $("#date_employed").val();
var gender = $("#sex").val();
//console.log(txt);
if(txt == '' && drop == "choose" && gender == "specify")
{
$("#txtSearch").css('border-color', 'red');
}
else
{
if(drop == "choose" && gender == "specify")
{
$.ajax
({
url: 'search.php',
type: 'POST',
data: {data1: txt, date_emp: drop, gen: gender},
dataType: 'JSON',
success:function(res)
{
$("#myTable tr").remove();
$("#myTable").append("<tr><th>Name</th><th>Position</th><th>Salary</th><th>Date</th><th>Gender</th></tr>");
$.each( res, function(key, row){
$("#myTable").append("<tr><td>"+row['emp_name']+"</td><td>"+row['position']+"</td><td>"+row['salary']+"</td><td>"+row['date_employed']+"</td><td>"+row['gender']+"</td></tr>");
});
},
error:function(res)
{
alert("Something Wrong");
}
});
}
$("#date_employed, #sex").change(function()
{
var txt = $("#txtSearch").val();
var drop = $("#date_employed").val();
var gender = $("#sex").val();
$.ajax({
url: 'search.php',
type: 'post',
data: {data1: txt, date_emp: drop, gen: gender},
datatype: 'json',
success:function(res)
{
$("#myTable tr").remove();
$("#myTable").append("<tr><th>Name</th><th>Position</th><th>Salary</th><th>Date</th><th>Gender</th></tr>");
$.each( res, function(key, row){
$("#myTable").append("<tr><td>"+row['emp_name']+"</td><td>"+row['position']+"</td><td>"+row['salary']+"</td><td>"+row['date_employed']+"</td><td>"+row['gender']+"</td></tr>");
});
},
error:function(res)
{
alert("Couldn't find any data!");
}
});
});
}
});
WHERE gender and drop are 2 drop lists that forming a search filters
When I change one of the drop lists, per example, when I choose the date equal to: this week I should see in table 2 rows.
But I can only see them in the network (in devTool), and at console tab I see the following error:
Uncaught TypeError: Cannot use 'in' operator to search for 'length' in
[{"id":"48","0":"48","emp_name":"Alexa","1":"Alexa","position":"Secretary","2":"Secretary","salary":"8000","3":"8000","date_employed":"2016-02-23","4":"2016-02-23","gender":"female","5":"female"}]
The PHP code when I change drop lists is:
if($date_emp == "week" && $gender == "specify")
{
$search = "SELECT * FROM employee WHERE (emp_name = :val OR position = :val
OR salary = :val OR date_employed = :val
OR gender = :val) AND date_employed > DATE_SUB(NOW(), INTERVAL 1 WEEK)";
$searchStmt = $conn->prepare($search);
$searchStmt->bindValue(":val", $val);
$searchStmt->execute();
$res = $searchStmt->fetchAll();
echo json_encode($res);
}
When you make an ajax call and expect the response to be a json you need to send a json header from the PHP
header('Content-Type: application/json');
echo json_encode($data);
Sending the json header from the PHP will turn the "res" param in your ajax to a json object and not a json string.
If you don't send the the header you need to parse the json string into a json object
var json = JSON.parse(res);
I want to increment $j in $.each function of jquery. But $j is not getting incremented. What may be the reason. My code is as follows:
$.ajax
({
url : "test.php",
type : "post",
data : {
"categoryIds" : categoryIds
},
dataType : "json",
success : function(resp){
var html = "";
var i = 1;
<?php $j = 1; ?>
$.each(resp,function(key,questions ){
<?php
$j = 1;
$testdataIndex = "answer_".$j;
?>
var test = "<?php echo $testdataIndex?>";
alert(test);
var res = "<?php echo $_SESSION['testdata'][$testdataIndex]; ?> ";
var index = "<?php echo $j++;?>";
alert(index);
});
});
Each time test prints 'answer_1' and index as 1
Fixed session issue :
$.ajax
({
url : "test.php",
type : "post",
data : {
"categoryIds" : categoryIds
},
dataType : "json",
success : function(resp){
var html = "";
var i = 1;
<?php $j = 1; ?>
$.each(resp,function(key,questions ){
var testdataIndex = "answer_" + i ;
var filename = "getsession.php?sessionName="+testdataIndex;
$.get(filename, function (data)
{
sessionValues = data;
$("#"+testdataIndex).text(sessionValues);
});//get
html += '<textarea class="form-control answer" rows = "5" style="width:127%;" id="answer_'+i+'" name="answer_'+i+'" required="required"></textarea>';
i++;
});//each
});//ajax
getsession.php
<?php
session_start();
$sessionName = $_GET['sessionName'];
print ($_SESSION['testdata'][$sessionName]);
?>
I have answered my question for other who face same problem.
Hello I am attempting to create an ajax query but when my results are returned I get Undefined as a response. Except for the object called "hello" which returns back as "h" even though it is set to "hello". I have a feeling it has something to do with the way ajax is sending the data but i'm lost as to what may be the issue. Any help would be greatly appreciated.
here is the ajax
function doSearch() {
var emailSearchText = $('#email').val();
var keyCardSearchText = $('#keyCard').val();
var userNameSearchText = $('#userName').val();
var pinSearchText = $('#pin').val();
var passwordSearchText = $('#password').val();
$.ajax({
url: 'process.php',
type: 'POST',
data: {
"hello": "hello",
"emailtext": "emailSearchText",
"keycardtext": "keyCardSearchText",
"usernametext": "userNameSearchText",
"pinText": "pinSearchText",
"passwordtext": "passwordSearchText"
},
dataType: "json",
success: function (data) {
alert(data.msg);
var mydata = data.data_db;
alert(mydata[0]);
}
});
}
Then here is the php
include_once('connection.php');
if(isset($_POST['hello'])) {
$hello = $_POST['hello'];
$emailSearchText = mysql_real_escape_string($_POST['emailSearchText']);
$keyCardSearchText = mysql_real_escape_string($_POST['keyCardSearchText']);
$userNameSearchText = mysql_real_escape_string($_POST['userNameSearchText']);
$pinSearchText = mysql_real_escape_string($_POST['pinSearchText']);
$passwordSearchText = mysql_real_escape_string($_POST['passwordSearchText']);
$query = "SELECT * FROM Students WHERE (`User name`='$userNameSearchText' OR `Email`='$emailSearchText' OR `Key Card`='$keyCardSearchText')AND(`Password`='$passwordSearchText'OR `Pin`='$pinSearchText')";
$students = mysql_query($query);
$count = (int) mysql_num_rows($students);
$data = array();
while($student = mysql_fetch_assoc($students)) {
$data[0] = $student['First Name'];
$data[1] = $student['Last Name'];
$data[2] = $student['Date of last class'];
$data[3] = $student['Time of last class'];
$data[4] = $student['Teacher of last class'];
$data[5] = $student['Membership Type'];
$data[6] = $student['Membership Expiration Date'];
$data[7] = $student['Free Vouchers'];
$data[8] = $student['Classes Attended'];
$data[9] = $student['Classes From Pack Remaining'];
$data[10] = $student['5 Class Packs Purchased'];
$data[11] = $student['10 Class Packs Purchased'];
$data[12] = $student['Basic Memberships Purchased'];
$data[13] = $student['Unlimited Memberships Purchased'];
$data[14] = $student['Groupon Purchased'];
};
echo json_encode(array("data_db"=>$data, "msg" => "Ajax connected. The students table consist ".$count." rows data", "success" => true));
};
Your PHP script is likely producing error messages because the $_POST values you are trying to access don't match the key names you are sending in the request. For example: $_POST['emailSearchText'], yet you used emailtext in the AJAX call.
This is most likely causing jQuery to not be able to parse the response as JSON, hence the Undefined.
First of all, you have to remove the quotes or you will be passing those literals instead of the variables.
$.ajax({
...
data: {
hello: "hello",
emailtext: emailSearchText,
keycardtext: keyCardSearchText,
usernametext: userNameSearchText,
pinText: pinSearchText,
passwordtext: passwordSearchText
},
...
});
And then, like ashicus point out, in your PHP file:
$emailSearchText = mysql_real_escape_string($_POST['emailtext']);
$keyCardSearchText = mysql_real_escape_string($_POST['keycardtext']);
$userNameSearchText = mysql_real_escape_string($_POST['usernametext']);
$pinSearchText = mysql_real_escape_string($_POST['pinText']);
$passwordSearchText = mysql_real_escape_string($_POST['passwordtext']);
JS file looks ok, so this few clues to check PHP file.
See if there are even POST params there (printr all $_POST in php)
Check if your even enters IF. Add an else statement and echo some dummy json.
Check what is actually in encoded json that is echoed. Assign it to var then print.
I'm trying to pass image coordinates via json to a php file that should update the database.
I have this in jquery:
var coords=[];
var coord = $(this).position();
var item = { coordTop: coord.top, coordLeft: coord.left };
coords.push(item);
var order = { coords: coords };
$.ajax({
type: "POST",
data : +$.toJSON(order),
url: "updatecoords.php",
success: function(response){
$("#respond").html('<div class="success">X and Y Coordinates
Saved</div>').hide().fadeIn(1000);
setTimeout(function(){
$('#respond').fadeOut(1000);
}, 2000);
}
});
This is what updatecoords.php looks like:
<?php
$db = Loader::db();
$data = json_decode($_POST['data']);
foreach($data->coords as $item) {
$coord_X = preg_replace('/[^\d\s]/', '', $item->coordTop);
$coord_Y = preg_replace('/[^\d\s]/', '', $item->coordLeft);
$x_coord = mysql_real_escape_string($coord_X);
$y_coord = mysql_real_escape_string($coord_Y);
$db->Execute("UPDATE coords SET x_pos = '$x_coord', y_pos = '$y_coord'");
}
?>
The success message shows from the javascript however nothing gets updated in the database?
Any thoughts?
You have a + before $.toJSON. Meaning that the json string it returns will be converted to an integer - probably 0 or NaN.
Alex is right, but you can also take the quotes out from the preg_replace pattern. PHP naturally uses /.../ slashes to "quote" a regular expression.
try this
UPDATE coords SET x_pos = '".$x_coord."', y_pos = '".$y_coord."'
and look here how to use toJSON
EDIT :
try this
$.ajax({
type: "POST",
data : { coordTop: coord.top, coordLeft: coord.left ,coords: coords },
and in second code do this
$data = json_decode($_POST['data']);
$coordTop = mysql_real_escape_string($_POST['coordTop']);
$coordLeft = mysql_real_escape_string($_POST['coordLeft']);
$coords = mysql_real_escape_string($_POST['coords']);
foreach(.........
.....
.....
and then use those variables $coordTop , $coordLeft , $coords
Thanks for the help but this seemed to do the trick....
In javascript:
$.post('updatecoords.php', 'data='+$.toJSON(order), function(response){ alert (response + mydata);
In Updatecoords.php:
$data = json_decode(stripcslashes($_POST['data']));