Replace values of an array with another array in php - php

Question

So, i am trying to display a chart where the profit and payout is rendering on a chart. If there is a no profit and payout value in chart then it should show 0.
I have an array with values of profit and payout with hour. Now i want to replace that array with an existing array of 0 values.
Here is my code
$a1=array();
for($i=0,$i<=24,$i++){
$a1['hour']=$i;
$a1['payout']='0';
$a1['profit']='0';
}
$a2=array();
$a2['hour']='2';
$a2['profit']='300';
$a2['payout']='100';
print_r(array_replace($a1,$a2));
There is something wrong with this code. Can any1 tell me what i am doing wrong?

Answer 1

<?php
// Initial array
$a1=array();
for($i=0;$i<=24;$i++){
// Use hour as index of array, if you use $a2[] = array(), it works
// But problem is when you change hours, lets say 12-24, if will cause problem
$a1[$i] = array(
'hour'=> $i,
'payout'=> 0,
'profit'=> 0
);
}
// Array from database
$a2=array();
$a2[] = array(
'hour'=> 2,
'payout'=> 300,
'profit'=> 100
);
$a2[] = array(
'hour'=> 5,
'payout'=> 3500,
'profit'=> 1200
);
echo '<pre>';
// Loop through second array and check if it is there in first one.
foreach( $a2 as $item) {
if(isset($a1[$item['hour']])) {
// Replace the values
$a1[$item['hour']] = $item;
}
}
print_r($a1);
?>
You are using for loop in wrong way, SyntaxError
for($i=0;$i<=24;$i++){ // <= See semi colons
}

Answer 2

First of all your For Loop isn't correct ! you must replace the "," with ";"

Answer 3

You are having a syntax error in your program.
Your working program should look something like this
$a1=array();
for($i=0;$i<=24;$i++){
$a1['hour']=$i;
$a1['payout']='0';
$a1['profit']='0';
}
$a2=array();
$a2['hour']='2';
$a2['profit']='300';
$a2['payout']='100';
print_r(array_replace($a1,$a2));
Hope this helps!