Why I can't run this query?

Question

I'm trying to run this code in php but does not print anything.

My database is not empty.

<?php
$dbname = "mytest";
$servername = "***";
$username = "mohsen";
$passwords = "****";
// Create connection
$conn = new mysqli($servername, $username, $passwords, $dbname);
// Check connection
if ($conn->connect_error) {
 die("Connection failed: " . $conn->connect_error);
}
$User=$_POST['username'];
$sql="SELECT * FROM Money WHERE Debtor='$User' OR Creditor='$User'  ";
$result = $conn->query($sql);

//
while($field=mysql_fetch_assoc($result))
{

print $field["Debtor"]."|";
print $field["Creditor"]."|";
print $field["Cost"]."|";
print $field["Status"]."|"."\n";

}
$conn->close();
?>

EX: I have record with this user in my database User=mohsen link:

http://test.kholaseketab.ir/Update.php

`mysql_fetch_assoc($result)` should be `$result->fetch_assoc()` - you are mixing APIs. — Qirel, Jan 30 '16 at 10:20
You need change `while($field=$result->fetch_assoc())` instead of `while($field=mysql_fetch_assoc($result))`. — Hardik Solanki, Jan 30 '16 at 10:20
@hardiksolanki Minor typo there... It's not `$conn->fetch_assoc($result)`, you seem to be mixing procedural and objectoriented. It should be `$result->fetch_assoc()` — Qirel, Jan 30 '16 at 10:21

score 1 · Answer 1 · answered Jan 30 '16 at 10:23

You are mixing object oriented and procedural style query.

<?php
$dbname = "mytest";
$servername = "***";
$username = "mohsen";
$passwords = "****";
// Create connection
$conn = new mysqli($servername, $username, $passwords, $dbname);
// Check connection
if ($conn->connect_error) {
 die("Connection failed: " . $conn->connect_error);
}
$User=$_POST['username'];
$sql="SELECT * FROM Money WHERE Debtor='$User' OR Creditor='$User'  ";
$result = $conn->query($sql);

//
while($field = $result->fetch_object())
{

print $field->Debtor."|";
print $field->Creditor."|";
print $field->Cost."|";
print $field->Status."|"."\n";

}
$conn->close();
?>

Well, kinda.. He's mixing APIs, `mysql_` with object oriented MySQLi. Which means that it's a dupe. — Qirel, Jan 30 '16 at 10:24

score 0 · Answer 2 · edited Jan 30 '16 at 11:35

<?php
   $dbname = "mytest";
   $servername = "***";
   $username = "mohsen";
   $passwords = "****";

   // Create connection
  $conn = new mysqli($servername, $username, $passwords, $dbname);

  // Check connection
  if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
  }

  $User=$_POST['username'];
  $sql="SELECT * FROM Money WHERE Debtor='$User' OR Creditor='$User'  ";
  $result = $conn->query($sql);

  while($field=$result->fetch_assoc())
  {
     print $field["Debtor"]."|";
     print $field["Creditor"]."|";
     print $field["Cost"]."|";
     print $field["Status"]."|"."\n";
  }

  $conn->close();
?>

Why I can't run this query?

2 Answers2