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PHP MySql Query shows old value after update

I update the value in the sql table and then print the value, it still shows the old value.

<?
print_r(mysql_fetch_array(mysql_query("select visits from Orders")));
mysql_query("update Orders set visits=visits+1");
print_r(mysql_fetch_array(mysql_query("select visits from Orders")));
?>

It outputs 1, and then again 1. The second value should be 2. When I check in PhpMyAdmin it is 2, then why does it show Old value?

Please, help! Thanks in advance.

Regards, Mike

Edit:

This is the code that the OP tried to execute:

mysql_connect("localhost","mayankx_tt","111111");
mysql_select_db("mayankx_tt") or die(mysql_error()); 
print_r(mysql_fetch_array(mysql_query("select visits from Orders")));
mysql_query("update Orders set visits=visits+1");
print_r(mysql_fetch_array(mysql_query("select visits from Orders")));

And his conclusion:

Array ( [0] => 4 [visits] => 4 ) Array ( [0] => 4 [visits] => 4 )
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5 answers

Check if UPDATE is actually running:

mysql_connect("localhost","mayankx_tt","111111");
mysql_select_db("mayankx_tt") or die(mysql_error());

# Checking PRE Value
print_r(mysql_fetch_array(mysql_query("select visits from Orders")));

# Attempting to Modify
if( !mysql_query("update Orders set visits=visits+1") ){
  echo 'Error Occurred: #'.mysql_errno().' '.mysql_error();
}else{
  echo 'UPDATE Processed OK';
}

# Check POST Value
print_r(mysql_fetch_array(mysql_query("select visits from Orders")));

Run this and let us (me) know what result it produced.

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, . ? CLI - phpmyadmin/sequel pro root. . , , /.

0

, , Orders. WHERE UPDATE SELECT, /row .

0

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SELECT * FROM TABLE_NAME 1

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You may receive an error due to the use of orders as the tab name. "Orders . " This is a reserved keyword to use as a tab name for selection and updating. Attach it to the backtick (`) symbol

"Update `Orders` set visit = visit + 1"

To execute a Mysql query, it is better to track the error. As a rule, we register the sql instruction and register or display an error message in accordance with the setting. For simple debugging, use the following stmt command:

mysql_query( "SQL STMT" ) or die("MYSQL Error :: ".mysql_errno()."<br>Message: ".mysql_error());
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