PHP Javascript Array

Question

Hi I'm a novice PHP developer to say the least and I'm pretty stuck.

I'm trying to get the below code working I know I need a foreach loop or similar but I'm way out of my depth, I'm really not sure how to get this working, I know to most of you this is basic stuff but I'm lost.

Basically this is a receipt when a tab is settled. It searches the MySql Database for the relevant items and should print them on the receipt.

I know the MySql Query is not linked to the output at the bottom but I don't know how to do it.

<html lang="en">
 <head>
  <title>Receipt</title>
 </head>

 <body>
  <?php

   $invoicenum = $_POST['invoicenum'];
   $name = $_POST['name'];

   $netrev=$invoicenum - 1;

  ?>
  Items:
  <?
   $itemQuery = mysql_query("SELECT * FROM sales WHERE invoicenum = '$netrev' AND tabname = '$name'");
 $result = array();

 while($row = mysql_fetch_array($itemQuery))
 {
  $result[] = $row['itemname'];
 }

 echo json_encode($result);

   $amounts = json_decode($result['amounts']);
   $items = json_decode($result['items']);
   $prices = json_decode($result['prices']);
  ?>
   <br>
   <?
    for ($i = 0; $i < count($items); $i++)
    {
     echo $amounts[$i] . "x " . $items[$i] . " - " . $prices[$i] . "<br>";
    }
   ?>

 </body>
</html>

I've removed code that isn't relevant to this array, if you can help I'd be forever grateful.

I'm unsure what you question is... What is the current output? What is the desired output? What is working? What isn't working as intended? — ʰᵈˑ, Jul 28 '16 at 13:27
if `$result[]=$row['itemname']` why should `json_decode($result['amounts'])` work at all? — JustOnUnderMillions, Jul 28 '16 at 13:28
`mysql_query("SELECT * FROM sales WHERE invoicenum = '$netrev' AND tabname = '$name'");` **WARNING WARNING - Danger Will Robinson!** : http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php and http://bobby-tables.com/ — CD001, Jul 28 '16 at 13:30
This `$result[]=` creates number indexes , so no string key is present in the array — JustOnUnderMillions, Jul 28 '16 at 13:30
This isn't related to your question, but you should definitely look into mysqli and prepared statements, your SQL query is very vulnerable to SQL injection attacks. — Brandon Horsley, Jul 28 '16 at 13:31
Yes, mysql_ statements have been gone for 2 years. Please look into mysqli. As for your question, please elaborate. It is unclear what the desired output is and where your problem is. — TheValyreanGroup, Jul 28 '16 at 13:34

score 0 · Accepted Answer · answered Jul 28 '16 at 13:35

no need of encode and decode.

$itemQuery = mysql_query("SELECT * FROM sales WHERE invoicenum = '$netrev' AND tabname = '$name'");

while($row = mysql_fetch_array($itemQuery))
{
    echo $row['amounts'] . "x " . $row['items'] . " - " . $row['prices'] . "<br>";
}
 ?>

score 0 · Answer 2 · answered Jul 28 '16 at 13:35

0

Forget all json_decode/encode after filling $result then:

First do:

$result[] = $row;

Then:

foreach ($result as $set){
   echo $set['amount'] . "x " . $set['itemname'] . " - " . $set['price'] . "<br>";
   #do var_dump($set); here for checking the keys
}

answered Jul 28 '16 at 13:35

JustOnUnderMillions

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PHP Javascript Array

2 Answers2