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I am exporting data from table convert to CSV, to JavaScript variable. . But getting Error "fputcsv() expects parameter 2 to be array" . How to do it?

This is my PHP code
<?php
require_once 'd:\xampp\htdocs\bpr\pdo_db.php';
$sql = "Select ID,FULL_NAMe,AGE from TRIAL";
$csv = get_csv_string($dbh, $sql);
echo "<script> var_csv='" . $csv . "' </script>";
function get_csv_string($dbh, $sql)
{
try
{
$result = $dbh->query($sql);
//return only the first row (we only need field names)
$row = $result->fetch(PDO::FETCH_ASSOC);
if ($row == null) {
return "No Data";
}
$f = fopen('php://memory', 'r+');
foreach ($row as $field => $value) {
if (fputcsv($f, $field) === false) {
return false;
}
}
//second query gets the data
$data = $dbh->query($sql);
$data->setFetchMode(PDO::FETCH_ASSOC);
foreach ($data as $row) {
if (fputcsv($f, $row) === false) {
return false;
}
} // end record loop
rewind($f);
$csv = stream_get_contents($f);
return rtrim($csv);
} catch (PDOExepction $e) {
return $e->getMessage();
}
return $csv;
}
I am getting following error!
Instead of fputcsv(), I made my own code. But issue if found when I save the CSV to a javascript variable. Is there any other way to convert the data to CSV and pass it on to javascript?
I will convert this csv to JSON at Client side using javascript

You're trying to output a header row to CSV with the field names.
This code fails because you're looping through the $row array and attempting to write one field header at a time.
foreach ($row as $field => $value) {
if (fputcsv($f, $field) === false) {
return false;
}
}
What you want is an array with the field names in it. You can use array_keys() for that, so your code becomes
if (fputcsv($f, array_keys($row)) === false) return false;
No loop required.

ajax not work and not show php data

when one record then show data when multiple record come then not show data other site.
ajaxx.php
<?php
include 'database.php';
session_start();
$post = $_POST;
$search = $post['search'];
$searchType = $post['searchType'];
if ($searchType == 'all')
{$sql = "SELECT DISTINCT title FROM hadees WHERE title LIKE '$search%' AND (type='Bukhari' OR type='Muslim') ";}
else
{$sql = "SELECT DISTINCT title FROM hadees WHERE title LIKE '$search%' AND type='$searchType' ";}
$result = mysqli_query($db,$sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$row['title'];
echo json_encode($row);
}
} else
{ echo "Not Found Result" ; }
?>
when data record is one then append the data successfully when multiple record come then not show data and append not work
javascript code
function searchh()
{
var type = $("input[name='type']:checked").val();
var searchhh = $( ".myButton option:selected" ).text();
debugger;
$.ajax({
url: 'ajaxx.php',
type: "POST",
data: {'searchType':type, 'search':searchhh},
success: function (data) {
var duce = jQuery.parseJSON(data);
alert(duce.title);
}
});
}

I think your issue is in the while loop. You don't want to encode each row one-by-one, but as a whole like this.
$myResults = [];
while($row = $result->fetch_assoc()) {
$row['title'];
$myResults[] = $row;
}
echo json_encode($myResults);

You are producing invalid JSON by using echo json_encode($row); within a loop.
Try to craft an array of rows, and then display it.
if($result->num_rows > 0)
{
$output = array();
while($row = $result->fetch_assoc())
{
output[] = $row;
}
if($searchType == 'all')
{
echo json_encode($output);
}
else
{
echo json_encode(current($output)); // print just one
}
}

Passing a 2D Array from PHP to Javascript

I'm a beginner in using PHP and Javascript, and I don't have any idea on how to store the data that I've gathered from MySQL which I placed in a multidimensional array in PHP to a 2D array in Javascript. Here's my working code in PHP:
<?php
function connecToDatabase(){
$host = "localhost";
$username = "root";
$password = "p#ssword";
$database = "flood_reports";
mysql_connect("$host", "$username", "$password") or die(mysql_error());
mysql_select_db("$database") or die(mysql_error());
}
function retrieveData(){
connecToDatabase();
$data = mysql_query('SELECT * FROM entries') or die(mysql_error());
$entries = array();
$index = 0;
while($info = mysql_fetch_array( $data ))
{
$entries[$index] = array('entry_id' => $info['entry_id'],
'location' => $info['location'],
'image_dir' => $info['image_dir'],
'longitude' => $info['longitude'],
'latitude' => $info['latitude'],
'level' => $info['level']);
$index++;
}
$json = json_encode($entries);
echo $json;
mysql_close();
}
retrieveData();
?>

on the end of your script add the following
<script type="text/javascript">
var jsvar = <?php echo $phpvar;?>
</script>

Replace
echo $json;
with
echo 'var fromPhp = ' . $json . ';';
You just need to put the data into a variable. This will make it available as fromPhp on the browser side.

Getting value from file called by ajax request using json

This file is called by ajax request. And result coming here I want to place into two different in calling function.
<?php
//Some processing gives $text
$s=nl2br($text);
$data['x'] = $p;
$data['y'] = $q;
//Start from here
echo "<b>Positive count : $x with $p % </b>"; echo "</br>";
echo "<b>Negative count : $y with $q % </b>"; echo "</br>";
echo "</br>";
echo "Page content : ";
echo "</br>";
echo "</br>";
echo $s;
//End. This content should be place in <div1>. Want to send this as a json string
and
//Start from here
echo "First 5 post";
$result = mysqli_query($con,"select post from facebook_posts where p_id > (select MAX(p_id) - 5 from facebook_posts)");
while ($row = $result->fetch_array(MYSQLI_ASSOC))
{
echo $row['post'];
echo '<br/>';
}
//End. This content should be placed in <div2> Want to send this as a json string
If there is single variable then we can easily do it using :
$resultArray = array("resultOne" => $result1,"resultTwo" => $result2);
echo json_encode($resultArray);
at receiving end:
document.getElementById("myFirstDiv").innerHTML=xmlhttp.responseText.resultOne;
document.getElementById("mySecondDiv").innerHTML=xmlhttp.responseText.resultTwo;
But how above complex result could be place into to json variable?

You could use output buffering in PHP:
ob_start();
// Generate content for div 1...
$div1 = ob_get_clean();
ob_start();
// Generate content for div 2...
$div2 = ob_get_clean();
$result = array("div1" => $div1, "div2" => $div2);
echo json_encode($result);

Display All Data with PHP AJAX JSON using JQUERY

filePHP.php
$query = $kon->prepare("SELECT * FROM t_kategori");
$query->execute();
while($row = $query->fetch(PDO::FETCH_ASSOC))
{
$json = array('id' => $row['id_kategori'], 'nama' => $row['nama_kategori']);
echo json_encode($json);
}
and index.php
$.post('filePHP.php', function(data){
console.log(data);
},'json');
but this not working, what would solve this problem?

Try this in PHP
$query = $kon->prepare("SELECT id_kategori,nama_kategori FROM t_kategori");
$query->execute();
$json=array();
while($row = $query->fetch(PDO::FETCH_ASSOC))
{
$arr=array('id'=>$row['id_kategori'],'nama'=>$row['nama_kategori']);
array_push($json,$arr);
}
echo json_encode($json);
Read array-push

try something like this , your code will echo json in wrong format whereas below code will give you json array.
$query = $kon->prepare("SELECT * FROM t_kategori");
$query->execute();
$json_arr =array();
while($row = $query->fetch(PDO::FETCH_ASSOC))
{
$temp_arr = array();
$temp_arr['id'] => $row['id_kategori'];
$temp_arr['nama'] => $row['nama_kategori'];
array_push($json_arr,$temp_arr);
}
echo json_encode($json_arr);