Logan Voss Logan Voss - 1 year ago 54
PHP Question

How to get number of results in this object?

I have an ajax post that turns the JSON that looks like this

{"0":{"Project_ID":"1","Project_Name":"TEST 1"...},

into an object that can be accessed by using
result[0]['Whatever Key']
result[1]['Whatever Key']
. Most of the data is the same considering my query grabs information for just 1 project, but the reason I have multiple results is I have data that fits into more than one location, or cost center category, etc.

I'm unsure how to iterate through the number of results and use that number in a for loop to populate a table with all the values instead of just the ones in index

The script looks like this:

echo "<script type'text/javascript'>
$('#result_year').change(function(event) {
$.post('info.php', { selected_project: $('#selection_project option:selected').val(), selected_year: document.getElementById('result_year').value},
function(data) {
var result = JSON.parse(data);
//maybe an $.each loop here? or
//for(var i=0; i<result.length; i++) <--(but NOT the length, the number of rows?)
var location = result['Location'];
var cccName = result['Cost_Center_Category_Name'];
var ccName = result['Cost_Center_Name'];
var Q_1 = result['Q_1'];
var Q_2 = result['Q_2'];
var Q_3 = result['Q_3'];
var Q_4 = result['Q_4'];

Here's my info.php as well

require ('system.php');

if (isset($_POST['selected_project']))
$selected_project = $_POST['selected_project'];

if (isset($_POST['selected_year']))
$selected_year = $_POST['selected_year'];

$query = "SELECT Project_Info.Project_ID, Project_Name, Global_Project_Number, Project_Type.Project_Type_Name, Lead_Location, Local_Project_Number_1, Local_Project_Number_2, Local_Project_Number_3, Local_Project_Number_4, Local_Project_Number_5, Development_Location_2, Development_Location_3, Development_Location_4, Development_Location_5, Customer, Duration, Customer_Group, Average_Number_of_Pieces, State, Business_Center.BC_Name, Start_Of_Production, Outlets.Outlet_Name, End_Of_Development, Start_Of_Development, Project_Connection, Project_Manager, Chance_Budget, System_TPL, Chance_Forecast, Product_Family, Technology, LL_SW, Processor_Type, HL_SW, Chassis_Type, Technology_Development, Technical_Description, username, Milestones.Description AS Milestone_Description, Milestone_1, Milestone_2, Milestone_3, Milestone_4, Milestone_5, Milestone_6, Due_Date_1, Due_Date_2, Due_Date_3, Due_Date_4, Due_Date_5, Due_Date_6, Labour_Planning.*, Cost_Center_Category.Cost_Center_Category_Name, Cost_Centers.Cost_Center_Name
FROM Project_Info
LEFT JOIN Project_Type ON Project_Type.Project_Type_ID = Project_Info.Project_Type
LEFT JOIN Business_Center ON Business_Center.BC_ID = Project_Info.BU_CC
LEFT JOIN Outlets ON Outlets.Outlet_ID = Project_Info.Outlet
LEFT JOIN Milestones ON Milestones.Project_ID = Project_Info.Project_ID
LEFT JOIN Labour_Planning ON Labour_Planning.Project_ID = Project_Info.Project_ID
AND Labour_Planning.Year = $selected_year
LEFT JOIN Cost_Center_Category ON Cost_Center_Category.Cost_Center_Category_ID = Labour_Planning.Cost_Center_Category
LEFT JOIN Cost_Centers ON Cost_Centers.Cost_Center_Number = Labour_Planning.Cost_Center_Number
WHERE Project_Info.Project_ID = '$selected_project';";
$result = $mysqli->query($query);

while($row = $result->fetch_assoc()){
$results[] = $row;
$results['selected_year'] = $selected_year;
$results['selected_project'] = $selected_project;
$response = json_encode($results);

echo $response;


Any help would be very much appreciated!

Answer Source

The problem is that you're mixing your database records into the same PHP array as the metadata for the selected year and project. Put the records into a nested array, so you can loop over them.

while($row = $result->fetch_assoc()){
    $data[] = $row;
$results['selected_year'] = $selected_year;
$results['selected_project'] = $selected_project;
$results['data'] = $data;
$response = json_encode($results);

echo $response;

Then in the jQuery code you can use result.data.length to get the number of records, and loop over them with $.each(result.data, function(index, row) {... });

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