[SOLVED] Query wont update 3 records...

[SirChick]

I have a update query in this script below :

 

$get_users = mysql_query("SELECT * FROM userregistration WHERE CountryID='1' AND IllnessType='0'");
if (!$get_users)
    die("Error: " . mysql_error());

//Loop through all the records
while($user = mysql_fetch_array($get_users)) {

  //Do something which has a 20% chance of succeeding.
  //Choose a random number between 1 and 5
  $rand = rand(1,1);

  //If the number is 3 (which there is a 20% chance of it being) update the user
  if($rand == 1) {
    if (($q = mysql_query("UPDATE userregistration SET IllnessType='Winter Flu' And TimeOfIllness='$Date' WHERE CountryID='1'")) != FALSE)
  print "update successful...<br>";
else
  print mysql_error(); 
}

}

 

but it just won't do it for the life of it ...there is deffinately 3 records as the update succesful echo's 3 times but in my database the fields are not updating to the queries input :S why could this be ?

 

before you ask..i put rand to 100% to test it.

[papaface]

try:

 

$get_users = mysql_query("SELECT * FROM userregistration WHERE CountryID='1' AND IllnessType='0'");
if (!$get_users)
    die("Error: " . mysql_error());

//Loop through all the records
while($user = mysql_fetch_array($get_users)) {

  //Do something which has a 20% chance of succeeding.
  //Choose a random number between 1 and 5
  $rand = rand(1,1);

  //If the number is 3 (which there is a 20% chance of it being) update the user
  if($rand == 1) {

$q = mysql_query("UPDATE userregistration SET IllnessType='Winter Flu' And TimeOfIllness='$Date' WHERE CountryID='1'");
    if ($q)
{
  print "update successful...<br>";
}
else
{
  print mysql_error(); 
}
}

}

[SirChick]

update successful
update successful
update successful

 

thats the result.. but in my database IllnessType and TimeOfIllness still are "0" and "0000-00-00 00:00:00" on all 3 records.

[dbo]

What are the datatypes for the fields that aren't updating?

[SirChick]

IllnessType - VarChar i tried Text also but no such luck..

TimeOfIllness - timestamp

 

 

encase you wondered what $Date is:

$Date = date("Y-m-d H:i:s",time());

[SirChick]

Just to add.. i found one problem i had AND in my query which i changed to a , symbol..... though it fixed it and worked it then decided to stop working again for different unknown reasons... now the ending echo's neither of them echo so its not succesful or an error:

 

$get_users = mysql_query("SELECT * FROM userregistration WHERE CountryID='1' AND IllnessType='0'");
if (!$get_users)
    die("Error: " . mysql_error());

//Loop through all the records
while($user = mysql_fetch_array($get_users)) {

  //Do something which has a 20% chance of succeeding.
  //Choose a random number between 1 and 5
  $rand = rand(1,5);

  //If the number is 3 (which there is a 20% chance of it being) update the user
  if($rand == 3) {
$Date = date("Y-m-d H:i:s",time());
$q = mysql_query("UPDATE userregistration SET IllnessType='Winter Flu', TimeOfIllness='$Date' WHERE CountryID='1'");
    if ($q)
{
  print "update successful...<br>";
}
else
{
  print mysql_error(); 
}
}

}

[cooldude832]

do use if($q) use $r = mysql_query($q) or die(mysql_error());  also try echoing out the queries to see if they look right.

[SirChick]

the query when echo'd looked fine but i dont get how it did work but now it isn't even though i aint altered it ...

 

 

Current code:

$get_users = mysql_query("SELECT * FROM userregistration WHERE CountryID='1' AND IllnessType='0'");
if (!$get_users)
    die("Error: " . mysql_error());

//Loop through all the records
while($user = mysql_fetch_array($get_users)) {

  //Do something which has a 20% chance of succeeding.
  //Choose a random number between 1 and 5
  $rand = rand(1,5);

  //If the number is 3 (which there is a 20% chance of it being) update the user
  if($rand == 3) {
$Date = date("Y-m-d H:i:s",time());
$q = mysql_query("UPDATE userregistration SET IllnessType='Winter Flu', TimeOfIllness='$Date' WHERE CountryID='1'");
    $r = mysql_query($q) or die(mysql_error());
}

}

[cooldude832]

can I see the outputted queries/ or mysql errors if any

[SirChick]

Just echo'd it and this came out:

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1

 

This is the script at the moment:

 

$get_users = mysql_query("SELECT * FROM userregistration WHERE CountryID='1' AND IllnessType='0'");
if (!$get_users)
    die("Error: " . mysql_error());

//Loop through all the records
while($user = mysql_fetch_array($get_users)) {

  //Do something which has a 20% chance of succeeding.
  //Choose a random number between 1 and 5
  $rand = rand(1,5);

  //If the number is 3 (which there is a 20% chance of it being) update the user
  if($rand == 3) {
$Date = date("Y-m-d H:i:s",time());
$q = mysql_query("UPDATE userregistration SET IllnessType='Winter Flu', TimeOfIllness='$Date' WHERE CountryID='1'");
    $r = mysql_query($q) or die(mysql_error());
}
echo $q;
}

[only one]

You are setting two mysql_quers in variables, so it is not actually running any query:

 

mysql_query("UPDATE userregistration SET IllnessType='Winter Flu', TimeOfIllness='$Date' WHERE CountryID='1'") or die(mysql_error());