jquery autcomplete returning source array - php

I'm using jquery's ajax function to fetch data from an external php file. The data that is returned from the php file will be used for the autocomplete function. But, instead of the autocomplete function suggesting each particular value from the array in the php file, it returns ALL of them. My jquery looks like this.
jQuery('input[name=past_team]:radio').click(function(){
$('#shadow').fadeIn('slow');
$('#year').fadeIn('slow');
var year = $('#year').val();
$('#year').change(function () {
$('#shadow').val('');
$.ajax({
type: "POST",
url: "links.php",
data: ({
year: year,
type: "past_team"
}),
success: function(data)
{
var data = [data];
$("#shadow").autocomplete({
source: data
});
}
});
});
});
The link.php file looks like this:
<?php
session_start();
require_once("functions.php");
connect();
$type = $_POST['type'];
$year = $_POST['year'];
if($type == "past_team")
{
$funk = mysql_query("SELECT * FROM past_season_team_articles WHERE year = '".$year."'")or die(mysql_error());
$count = mysql_num_rows($funk);
$i = 0;
while($row = mysql_fetch_assoc($funk))
{
$name[$i] = $row['team'];
$i++;
}
$data = "";
for($i=0;$i<$count;$i++)
{
if($i != ($count-1))
{
$data .= '"'.$name[$i].'", ';
} else
{
$data .= '"'.$name[$i].'"';
}
}
echo $data;
}
?>
The autocomplete works. But, it's just that when I begin to enter something in the input field, the suggestion that are loaded is the entire array. I'll get "Chicago Cubs", "Boston Red Sox", "Atlanta Braves", .....

Use i.e. Json to render your output in the php script.
ATM it's not parsed by javascript only concaternated with "," to a single array element. I do not think that's what you want. Also pay attention to the required datastructure of data.
For a working example (on the Client Side see the Remote JSONP example http://jqueryui.com/demos/autocomplete/#remote-jsonp )

Related

Trigger combo box events using jquery

I have this code:
$show_location = mysql_query("SELECT * FROM location ORDER BY location_code");
while($row_location = mysql_fetch_array($show_location))
{
$location_code = $row_location['location_code'];
$show_store = mysql_query("SELECT * FROM store_list WHERE location LIKE '%$location_code%'");
$count_store = mysql_num_rows($show_store);
while($row_store = mysql_fetch_array($show_store))
{
$store_name = $row_store['store_name'];
}
if($count_store==0)
{
$status = "Inactive";
echo '<option value="'.$location_code.'">'.$location_code.'</option>';
$sql1 = "SELECT description FROM location WHERE location_code=$location_code";
$result1 = mysql_query("$sql1");
$row1 = mysql_fetch_assoc($result1);
$description=$row1['description'];
}
else
{
$status = "Active";
}
//echo '<option value="'.$location_code.'">'.$location_code.'</option>';
}
What I want to do is to display the $description somewhere in the form. I have the kind of combobox where you can select as many as you can. I want to display each $description once a location is selected. But I dont know where to put the trigger. Can sombody help me? Thanks!
correct me if i am wrong, but will every location_code have one description right?
well there are two ways:
1) Easy but not so efficient way
Make an ajax call for every selected value.
$("#myDropDown").change(function (event) {
//alert("You have Selected :: "+$(this).val());
$.ajax({
type: "POST",
url: "ajax.php",
data: { type: "1", location: $(this).val() }
}).done(function( msg ) {
("#mydata").html(msg)
});
});
You can check for type == 1 on php side, get the description and print it
The above method will cause a ajax request for every selection
2) A bit complex, but efficient
First of all json_encode your location_code and description. It will become something like {code:"AU", description:"blah blah"}
Then use something like this
$("#sel").change(function(e){
//alert($(this).val());
var array = $(this).val();
for(key in array){
var json = JSON.parse(array[key]);
$("#desiptions").html(json.description);
}
});

how to fetch the array data from database query in a php file to ajax sucess data

i am trying to get the data which is in array, from an SQL query to the ajax success data. Here is the code i tried,
function adminchecksub(value1){
//alert('hi');
var value1=$('#adminsid').val();
//alert(value1);
if(value1==''){
alert('Please enter a Sub id');
}
else{
//alert(value1);
$.ajax({
type: 'POST',
url: root_url + '/services/services.php?method=adminsubcheck',
data: {value1:value1},
async: true,
success: function (data) {
alert(data);
if (data == 1) {
alert('inside');
//window.location.assign(rdurl+"?sid="+sid);
return true;
} else {
//alert('does not exist!');
//alert('outside');
return false;
}
}
});
}
now in the php method which is requested from the above ajax call, i have to get the data return by this function which is returning an array
function adminsubcheck($sid){
$subid=$sid['value1'];
$row = array();
//echo $subid;
$query = "SELECT Sub_id,Status,Sub_type FROM Sub WHERE Sub_id=$subid";
//echo $query;
//echo $subid;
$queryresult = mysql_query($query);
$count = mysql_num_rows($queryresult);
//echo $count;
while ($r = mysql_fetch_assoc($queryresult)) {
$row[] = $r;
}
return $row;
}
here $row is returning an array as data to the ajax function, where i need to get all the items seperately. Some one help me out to get the values into the ajax call. Thanks in advance..
In your PHP file, you can return your array as a json object in the following way (notice how the PHP file echos the result to pass it back to javascript)
...
$json = json_encode($row);
echo $json
Now add the following to your javascrip ajax
$.ajax({
...
// Whatever code you currently have
...
}).done(function ( json ) {
var data = JSON.parse(json);
//Continue with javascript
});
You can handle the java script variable 'data' as an associative array like it was in PHP
also, look how you populate the php variable $row and adjust the following way:
$cnt = 0;
while ($r = mysql_fetch_assoc($queryresult)) {
$row[$cnt] = $r;
$cnt++;
}
$json = json_encode($row);
echo $json;
in javascript you can access your rows the following way in teh data variable:
var value = data[0]['field_name'];
in the above statement, 0 will correspond to the value of $cnt (i.e. mysql returned row number)
return $row;
replace like this
echo json_encode($row);
and add dataType='json' in ajax like this
dataType: 'json',
If you want get the value in ajax call i think it shoul be :
while ($r = mysql_fetch_assoc($queryresult)) {
$row[] = $r;
}
echo json_encode(array('data'=>$row));
exit;

Ajax variable not displaying on response

I am trying to incorporate some ajax into my code, at the moment it works great, when a user clicks a button it sends some hidden inputs to another php script, performs some actions, then sends the result back and outputs it, all updating with no refresh etc.
However, the last part I am trying to send and receive back is a variable that shows some html code, in other words the other variables that are being sent back and outputed are just numbers and letters, where as this one is actual div's, however it is not outputting it, I have tried it without the ajax, when the page first loads and it works great, but doesn't when i try it with the ajax, hopefully the code below will make more sense. The variable I want to be able to send back is $sl_output.
AJAX code on main page
<script>
$(document).ready(function (){
$(".add_detail_land_down").click(function(){
var hidden_count = $('input[name=addon_detail_hidden_count]').val();
var land_required = $('input[name=addon_hidden_land_required]').val();
var sl_array = $('input[name=addon_hidden_shopping_list_array]').val();
var button_tok = "land_down";
$.ajax({
type: "GET",
url: "addon_detail_calc.php",
data: { hidden_count: hidden_count, button_tok: button_tok, land_required: land_required, sl_array: sl_array },
dataType: "json",
success: function (data) {
$("#res_expected_gain").html(data.total_hidden);
$("#output").html(data.output);
$("#res_expected_profit").html(data.land_required);
$("#res_total_supply_time").html(data.test_time);
$("#land_selected_token").html(data.total_hidden);
$("#sl_output_div").html(data.sl_output);
}
});
return false;
})
});
Code in addon_detail_calc.php
<?php
// header('application/json');
$hidden_count = $_GET["hidden_count"];
$button_tok = $_GET["button_tok"];
$land_required = $_GET["land_required"];
$sl_array = $_GET["sl_array"];
$sl_output = "";
if($button_tok == "land_up"){
//MAIN CODE SHALL BE DONE HERE
$hidden_count = $hidden_count + 1;
$test_time = $hidden_count * 66;
$new_sl_array = array();
$final_sl_array = array();
foreach ($sl_array as $columnName => $columnData) {
if($columnName == "0"){
unset($sl_array[$columnName]);
}else{
$new_columnData = $columnData * 54;
$new_sl_array[$columnName] = $new_columnData;
}
}
foreach ($new_sl_array as $columnName => $columnData) {
$sl_output = "";
$sl_output = '<li class="add_detail_content_ele_wrap_ele"><div class="add_detail_content_ele_wrap_ele_header">'.$columnName.'</div><!--end add_detail_content_ele_wrap_ele_header--><div class="add_detail_content_ele_wrap_ele_pic"></div><!--end add_detail_content_ele_wrap_ele_pic--><div class="add_detail_content_ele_wrap_ele_amount">47 Required</div><!--end add_detail_content_ele_wrap_ele_amount--><div class="add_detail_content_ele_wrap_ele_user_amount">You Have 0</div><!--end add_detail_content_ele_wrap_ele_user_amount--></li>';
$final_sl_array[$columnName] = $sl_output;
}
$array_result = implode("", $final_sl_array);
}else{
$hidden_count = $hidden_count;
}else{
$hidden_count = $hidden_count;
}
$output = "";
$output = '<input type="hidden" name="addon_detail_hidden_count" id="addon_detail_hidden_count" class="addon_detail_hidden_count" value="'.$hidden_count.'" />';
include 'connect_to_mysql.php';
echo json_encode(array("total_hidden" => $hidden_count, "output" => $output, "land_required" => $hidden_count, "test_time" => $test_time, "sl_output" => $final_sl_array ));
?>
I guess the question is, can the html code variable being transported be viewed and re-outputed?
EDIT
I have updated my code above, it seems that the first problem, was the foreach loop wasn't handling the right data, i had to shuffle things around and add another array. Because of this i have to put every value in array final_sl_array into a single variable, i have tried to implode this, it worked when i ran it through normal php when the page opens, however, now when i try it through ajax, it is not working, and not returning any result, any ideas why it works for one and not the other?

is php json_encode & AJAX breaking my array?

I'm grabbing some database entries, creating a 2D array and then passing them to js with AJAX. But when I loop through the array in javascript, it's an "undefined" mess. The console log for dbArray works fine, so I know the PHP/AJAX is working. Not sure what I am doing wrong with the loop...
PHP ('load-words.php):
$query = mysql_query("
SELECT * FROM words
ORDER BY RAND()
LIMIT 50
") or die(mysql_error());
$dbArray = array();
while ($row = mysql_fetch_assoc($query)) {
$word_phrase = stripslashes($row['word_phrase']);
$description = stripslashes($row['description']);
// construct a 2D array containing each word and description
$dbArray[] = array($word_phrase,$description);
};
echo json_encode($dbArray);
Javascript:
$.ajax({
url: 'func/load-words.php',
success: function(dbArray) {
console.log(dbArray);
var items = "<ul>";
for (var i in dbArray) {
items += "<li><a href='#'><b>" + dbArray[i][0] + ' : ' + dbArray[i][1] + "</a></li>";
}
items += "</ul>";
div = $('#dbArray');
div.html(items);
}
});
I guess this is failing because jQuery is interpreting the AJAX response as a string, since your PHP is not outputting a JSON header and your AJAX is not stipulating JSON. This is easily tested:
$.ajax({
url: 'func/load-words.php',
success: function(dbArray) { alert(typeof dbArray); /* "string"? */ }
});
Try
$.ajax({
url: 'func/load-words.php',
dataType: 'json', //<-- now we explicitly expect JSON
success: function(dbArray) { alert(typeof dbArray); /* "object"? */ }
});

loop over the PHP data Returned to jquery AJAX Call

below is my $.ajax call to php
$(document).ready(function() {
$('ul.sub_menu a').click(function(e) {
e.preventDefault();
var txt = $(this).attr('href');
$.ajax({
type: "POST",
url: "thegamer.php",
data:{send_txt: txt},
success: function(data){
$('#container').fadeOut('8000', function (){
$('#container').html(data);
$('#container').fadeIn('8000');
});
}
});
});
});
my php code
if(mysql_num_rows($result) > 0){
//Fetch rows
while($row = mysql_fetch_array($result)){
echo $row['img'];
}
}
I m getting this output
images/man/caps/army-black.pngimages/man/caps/army-brown.pngimages/man/caps/army-grey.pngimages/man/caps/army-lthr.pngimages
these are basically image paths now how to loop over them in jquery and fit each image in image tag
any code will be useful
Plz Note I DONT NEED JSON
regards sajid
JSON is probably your best bet here. In PHP do something like this:
$ret = array();
while( $row = mysql_fetch_assoc( $result ) )
{
$ret[] = $row['img'];
}
echo json_encode( $ret );
This will output something like the following
["image1","image2","image3"]
jQuery has a function which can convert this information into a javascript array. So put this code in your success callback.
var result = jQuery.parseJSON( data );
alert( result[1] );
EDIT: A method which does not use JSON
In PHP place each image url on a separate line
echo $row['img'], "\n";
Then in javascript, split the response by the new line character
var result = data.split( "\n" );
simply change your php code:
`if(mysql_num_rows($result) > 0){
while($row = mysql_fetch_array($result)){
echo ""<img src='".$row['img']."' /><br />";
} }

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