How to put my JSON info into a jQuery Array - php

I have a project where I've created JSON data for Project Prices/Prices I've gotten this far and pretty much ran into a wall, any help at all would help! I've checked all over the web and on jQuery.getJSON but I wound up getting super confused.
$data = mysql_query("SELECT * FROM xxx")
or die(mysql_error());
$arr = array();
$rs = mysql_query("SELECT product, price FROM products");
while($obj = mysql_fetch_object($rs)) {
$arr[] = $obj;
}
echo '{"products":'.json_encode($arr).'}';
I need to get the product price and product name into this jquery script
$(document).ready(function() {
/*** CONSTANTS ***/
var KEY = 0;
var VALUE = 1;
/*** DEFINE DATA SETS ***/
var POINTS = [ ["$productA", $PRICE ], ["$productB", $PRICE], ["$productC", $PRICE], ["$productD", $PRICE], ["$productE", $PRICE], ["$productF", $PRICE] ];
var SHIPPING_COSTS = [ ["Pickup", 0], ["Next Day Delivery", 30], ["Same Day Print/Same Day Delivery", 65] ];
for (var i = 0; i < POINTS.length; i++) {
$("#quantity").append("<option value='" + POINTS[i][VALUE] + "'>" + POINTS[i][KEY] + "</option>");
}
for (var i = 0; i < SHIPPING_COSTS.length; i++) {
$("#shipping").append("<option value='" + SHIPPING_COSTS[i][VALUE] + "'>" + SHIPPING_COSTS[i][KEY] + "</option>");
}
$("select.autoUpdatePrice, input.autoUpdatePrice").bind("mousedown click change", function(event) {
Calculate();
});
Calculate();
});
function Calculate() {
var net = parseFloat($("#quantity").val());
/* Calculate the magical # by adding the form fields*/
var designFee = $("#abcDesignFee").attr("checked") ? $("#abcDesignFee").val() : 0.0;
var proofFee = $("#abcProofFee").attr("checked") ? $("#abcProofFee").val() : 0.0;
var MyPrice;
MyPrice = parseFloat( parseFloat(proofFee) + parseFloat(designFee) + net + parseFloat($("#shipping").val()));
$("#DumpHere").html("Your Price: $" + formatNumber(MyPrice));
$("#abcName").val($("#quantity").find(":selected").text() + " " + ProductNamePlural);
$("#abcPrice").val(MyPrice);
}

In your PHP script, can you just json_encode() your array of objects without wrapping it in the string? And instead encode the JSON object like so:
<?php
// your script ...
echo json_encode($arr);
This creates an array of JSON encoded objects:
[{"name":"item 1","price":4.99},{"name":"item 2","price":9.99}]
Make an AJAX request in your JS to query your PHP script, and use jQuery's $.each() and $.parseJSON() methods to iterate over the returned JSON data:
$.post('get_products.php', { data: foo }, function(json) {
$.each($.parseJSON(json), function(key, product) {
console.log(product.name + ' ' + product.price);
});
});
Hope this helps :)

When I was learning - I had exactly the same problem - but it got answered here
What I didn't realise at the time was that you can use object notation (which is the ON of json) to access the data you sent.
If you look at my question and the answer I selected, once you have sent the data back to javascriot you can access it easily. In my example it would be as straitforward as using data.catagory_desc in javascript to find the data that was encoded.

Related

handling multiple db rows with ajax and json

I have a mysql database, I am using ajax to query it via php, it works fine when only a single row is returned it errors out with this when multiple rows are returned.
"xhr=[object Object]
textStatus=parsererror
errorThrown= SyntaxError: Unexpected token { in JSON at position 221"
Here is the code I am using, any help would be much appreciated.
$('button').click(function(){
event.preventDefault();
var box = document.getElementById('machine');
var rdata;
var id= box.options[box.selectedIndex].text;
$.ajax({
url: 'machine_report.php',
data: 'machine=' + id,
dataType: 'json',
success: function(rdata)
{
var uid = rdata[0];
var date = rdata[1];
var time = rdata[2];
var machine =rdata[3];
var reps = rdata[4];
var sets = rdata[5];
var weight = rdata[6];
var settings = rdata[7];
$('#status').html("<b>id: </b>"+uid+"<b> date: </b>"+date + "<b> Machine: </b>" + machine + "<b> reps: </b>" + reps + "<b> sets: </b>" +
sets + "<b> weight: </b>" + weight + "<b> settings: </b>" + settings);
},
error: function(xhr,textStatus, errorThrown) {
$("#status").html("xhr=" + xhr + "textStatus=" + textStatus + "errorThrown= " + errorThrown);
}
});
});
PHP Code
<?php
require_once("connect.php");
$machine = $_GET['machine'];
$mysql="SELECT * FROM workouts WHERE machine LIKE '$machine' ORDER BY uid DESC";
$result=mysqli_query($con,$mysql);
if (!$result) {
printf("Error: %s\n", mysqli_error($con));
exit();
}
$rowCount = mysqli_affected_rows($con);
while($row = mysqli_fetch_array($result)){
$date = $row['date'];
$time = $row['time'];
$machine = $row['machine'];
$reps = $row['reps'];
$sets = $row['sets'];
$weight = $row['weight'];
$settings = $row['settings'];
echo json_encode($row);
}
?>
Your PHP code is outputting one term at time so it will give response in following way.
{name:"abc",age:21}{name:"pqr",age:12}
which is not an valid json if you are passing multiple JSON Objects then you have to create collection of it.(which is simply a array of objects) The valid json for above json will be
[{name:"abc",age:21}{name:"pqr",age:12}]
You can produce this by first catching all the database rows into one array and then pass this array to json_encode
while($row=mysqli_fetch($result)
$rows[] = $row;
echo json_encode($rows);
All the above stuff is to do at the server side.
Now let's look at the client side.
Your Javascript code look fine except for success function because it is written for handling only one object.
Server response returns the data as array of json objects so first you have to iterate the array and then write a code for single object something like this
success:function(data) {
for(var i in data) {
var Row = data[i] // it will contain your single object
//console.log function is used for debugging purpose only you can find more about it by googling.
console.log(Row.name) // you can access it's member like this
// do other stuff such as dom manipulation.
}
}
NOTE: you can also use JQuery's $.each to traverse an array.

Passing a JSON array from PHP to JQuery is not working

I am having trouble accesssing a jquery array which is getting JSON data from a PHP script. If I hard coded the array in jquery it worked fine. I checked using cosole.log. Both nproducts and products array giving the same values. Please note that nproduct has hard coded values where are product is getting from a PHP script. Can someone put my in the right direction. Thanks
here is the PHP Code
while ($row = oci_fetch_array($result,OCI_ASSOC)) {
$shop[$row['WH_DESCRIPTION']] = array(
'pic' => $row['WH_PIC'],
'price' => $row['WH_PRICE']
);
}
echo json_encode($shop);
here is the jquery. If I use nproduct then productsRendering function works fine but if I use product then it print containsValue for name and pic and undefined for price. It seems that the product array does not have any values in the productRendering function where as getJSON is returning values.
<script type="text/javascript">
var cart = (function ($) {
var productsOffset = 3, products = [],
nproducts = {
'Black T-shirt': {
pic: 'black-controller.png',
price: 15
},
'Green T-shirt': {
pic: 'green-kentucky.png',
price: 18
},
'Brown T-shirt': {
pic: 'brown-computer.png',
price: 25
}
};
$.getJSON('shop.php', function(data) {
products = data;
console.log(data); //showing values here
console.log(products); //showing values here
console.log(nproducts); //showing values here
});
function render() {
productsRendering();
};
function productsRendering() {
var catalog = $('#catalog'),
imageContainer = $('</div>'),
image, product, left = 0, top = 0, counter = 0;
console.log(products); //does not have values here
for (var name in products) {
product = products[name];
image = createProduct(name, product);
image.appendTo(catalog);
if (counter !== 0 && counter % 3 === 0) {
top += 147; // image.outerHeight() + productsOffset;
left = 0;
}
image.css({
left: left,
top: top
});
left += 127; // image.outerWidth() + productsOffset;
counter += 1;
}
$('.draggable-demo-product').jqxDragDrop({ dropTarget: $('#cart'), revert: true });
};
function createProduct(name, product) {
return $('<div class="draggable-demo-product jqx-rc-all">' +
'<div class="jqx-rc-t draggable-demo-product-header jqx-widget-header-' + theme + ' jqx-fill-state-normal-' + theme + '">' +
'<div class="draggable-demo-product-header-label"> ' + name + '</div></div>' +
'<div class="jqx-fill-state-normal-' + theme + ' draggable-demo-product-price">Price: <strong>$' + product.price + '</strong></div>' +
'<img src="images/t-shirts/' + product.pic + '" alt='
+ name + '" class="jqx-rc-b" />' +
'</div>');
};
function init() {
render();
};
return {
init: init
}
} ($));
$(document).ready(function () {
cart.init();
});
</script>
productsRendering() gets called before ajax request completes. Consider calling productsRendering() inside the ajax callback.
This is because the browser does not interpret the response body as JSON. Try setting Content-Type header in php before echoing response:
header('Content-Type', 'application/json');

can't set variable in JavaScript based on json id from php

I have a simple array from a php file (the reason for not using a json file is cause this info comes from a mysql database)
$array[0] = array('id' => 1,'price' => '325');
$array[1] = array('id' => 2,'price' => '486');
header('Content-type: application/json');
echo json_encode($array);
and it echos as:
[{"id":1,"price":"325"},{"id":2,"price":"486"}]
now what I want to do is take the id and add it to a variable called counterval so that JavaScript will read it as
counterval1 = 325;
counterval2 = 486;
but I can't seem to get it to read that way. Here is the script at the current moment.
$.getJSON('test.php',function(data) {
$.each(data, function(i) {
counterval + data[i].id = data[i].price;
console.log (counterval2);
});
$('#results').html(counterval2);
});
var counterval1 = 0;
var counterval2 = 0;
any help on this would be greatly appreciated.
you can't do that... but you can do this...
var counterval = [];
$.getJSON('test.php',function(data) {
$.each(data, function(i) {
counterval[data[i].id] = data[i].price;
console.log (counterval[2]);
});
$('#results').html(counterval[2]);
});
counterval[1] = 0;
counterval[2] = 0;
See my comment on your post. If you really want the vars to look like you explained:
eval("counterval" + data[i].id + " = data[i]..price");
alert(counterval1);

jquerymobile selectmenu not populating

I am trying to populate a selected menu when the page is created but there are no options that show.
$(document).ready(function() {
$.ajax('patientlist.php', function(data){
var html = '';
var len = data.length;
for (var i = 0; i< len; i++) {
html += '<option value="' + data[i].patient_id + '">' + data[i].patient_firstname + data[i].patient_lastname + '</option>';}
$('#patientselect').append(html);
});
});
my patientlist.php
$result = mysql_query("SELECT `patient_id`, `patient_firstname`, `patient_lastname` FROM `patients` WHERE `company_id` = " . $user_data['company_id'] . " ORDER BY `patient_firstname`");
while($row = mysql_fetch_assoc($result)) {
$data[] = $row;
echo json_encode( $data );
}
My result from the php page
[{"patient_id":"9","patient_firstname":"Adam","patient_lastname":"Steve"}] etc...
Really appreciate any help, been stuck on this for a week!
So, posting again.
First of all, you should put the echo json_encode( $data ); out of your while loop
while($row = mysql_fetch_assoc($result)) {
$data[] = $row;
}
echo json_encode( $data );
Second, your $ajax syntax isn't correct, change this to $.post and tell the $.post request you are expecting a 'json' response from patientlist.php
$(document).ready(function() {
$.post('patientlist.php', {}, function(data){
/* code */
}, 'json'); // <= set data type json here
});
When retrieving a valid json string, you can iterate over data by using the $.each method
$.each(data, function(index, patient) {
console.log(patient.patient_id); // or use patient['patient_id']
});
At least you will now receive a valid request.
Noticing your HTML, do not use .append if it is not a DOM element, you are just building html elements as a string, so use instead
$('#patientselect').html(html);

How do I list rows from a SQL table into my jquery script?

I'm creating a pricing array jQuery and for this project the amount of different products I need is going to be +200 so the easiest option I could research would be listing from a SQL table.
This is what I have so far:
$(document).ready(function() {
/*** CONSTANTS ***/
var KEY = 0;
var VALUE = 1;
/*** DEFINE DATA SETS ***/
var POINTS = [ ["250", 46.5 ], ["500", 53.5], ["1000", 67], ["2500", 107.5], ["5000", 175], ["10000", 310] ];
var SHIPPING_COSTS = [ ["Pickup", 0], ["Next Day Delivery", 30], ["Same Day Print/Same Day Delivery", 65] ];
for (var i = 0; i < POINTS.length; i++) {
$("#quantity").append("<option value='" + POINTS[i][VALUE] + "'>" + POINTS[i][KEY] + "</option>");
}
for (var i = 0; i < SHIPPING_COSTS.length; i++) {
$("#shipping").append("<option value='" + SHIPPING_COSTS[i][VALUE] + "'>" + SHIPPING_COSTS[i][KEY] + "</option>");
}
$("select.autoUpdatePrice, input.autoUpdatePrice").bind("mousedown click change", function(event) {
Calculate();
});
Calculate();
});
function Calculate() {
var net = parseFloat($("#quantity").val());
/* Calculate the magical # by adding the form fields*/
var designFee = $("#abcDesignFee").attr("checked") ? $("#abcDesignFee").val() : 0.0;
var proofFee = $("#abcProofFee").attr("checked") ? $("#abcProofFee").val() : 0.0;
var MyPrice;
MyPrice = parseFloat( parseFloat(proofFee) + parseFloat(designFee) + net + parseFloat($("#shipping").val()));
$("#DumpHere").html("Your Price: $" + formatNumber(MyPrice));
$("#abcName").val($("#quantity").find(":selected").text() + " " + ProductNamePlural);
$("#abcPrice").val(MyPrice);
}
Create a PHP file and read the data from SQL into an array. Print that array after json_encode(). Then use jQuery.getJSON() on document ready to fetch it and process it.

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