How to display multiple array value in PHP? - php

Question

I am having an trouble in displaying values in PHP.
Value pass through URL career.php?mode=1,2,3
Here is my code
$id = $_GET['mode']; // Id get from URL
//echo $id;
$query = "SELECT `job_title` FROM `job` WHERE `job_id` IN ($id)";
$res = mysqli_query($conn, $query);
foreach(($row = mysqli_fetch_array($res)) as $key1){
$key[] = $key1;
}
PHP Code:
<?php echo $key;?>
Hence it is in looping so it shows the last looped value. Is it possible to display the all values through loop.
Help me out guys!!

Answer 1

You have multiple options to output an array.
Echo and foreach
You c an loop through your array and echo each $key and $value
foreach ($keys as $key => $value) {
echo $key." : ".$value."<br />";
}
print_r and var_dump
This is mostly used in debugging.
print_r($keys);
var_dump($keys);
Imploding
You can implode your array to echo the concatted values.
// The first parameter is the devider or separator
echo implode('', $keys);
Is it possible to display more than one value from the column?
Yes, ofcourse. Look at the following example:
foreach ($keys as $value) {
echo $value['columnone'];
echo $value['columntwo'];
echo $value['columnthree'];
}
Resources
implode() - PHP Manual
print_r() - PHP Manual
var_dump() - PHP Manual
foreach - PHP Manual

Answer 2

As your code in open for sql injection you need to use bind and prepare statement . Use while loop to echo your data as
$ids[] = $_GET['mode']; // store it into array
$query = "SELECT `job_title` FROM `job` WHERE `job_id` IN (";
$query .= implode(',', array_fill(0, count($ids), '?'));// bind your param
$query .= ') ';
$stmt = $conn->prepare($query);
call_user_func_array(array($stmt, 'bind_param'), $ids);//Call a callback with an array of parameters
$stmt->execute();
$stmt->bind_result($job_title);// bind result
while ($stmt->fetch()) {// use wlile loop here
printf("%s\n", $job_title);//echo result
}

Answer 3

You can do something like this:
$mysqli = new mysqli("server", "user", "password", "db");
$id = $_GET['mode'];
$query = "SELECT `job_title` FROM `job` WHERE `job_id` IN ($id)";
if($res = $mysqli->query($query)){
while($obj = $result->fetch_object()){
echo $obj->job_title;
}
}
$res->close();

Answer 4

Try this:
<?
$servername = "localhost";
$username = "yourUSERNAME";
$password = "yourPASS";
$dbname = "yourDBNAME";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
exit();}
$id = $_GET['mode']; // Id get from URL
//echo $id;
$query = "SELECT `job_title` FROM `job` WHERE `job_id`=".$id."";
$result = $conn->query($sql);
while($row=$result->fetch_assoc()){
//do something...
}
$conn ->close();
?>
tip: put your connection on another connection.php file.
like:
require('Connect.php');

Answer 5

$cars = array
(
array("Volvo",22,18),
array("BMW",15,13),
array("Saab",5,2),
array("Land Rover",17,15)
);
for ($row = 0; $row < 4; $row++) {
echo "<p><b>Row number $row</b></p>";
echo "<ul>";
for ($col = 0; $col < 3; $col++) {
echo "<li>".$cars[$row][$col]."</li>";
}
echo "</ul>";
}