[Resolved] Checking if it exists

[Sharif]

This is the code I use:
<?php
$display_title = mysql_query("SELECT pagetitle FROM legal WHERE pageid=$pageid")
or die ("MySQL Error - Please report this to the webmaster");
if ($pageid < 1 OR $pageid >6)
{
echo("No Such Page");
}
else
{
while ($row = mysql_fetch_array($display_title))
{
$legal = $row["pagetitle"];
echo $legal;
}
}
?>

This code is used on a page, legal.php to show legal notices (privacy policy, terms of use, etc.). In the database where the legal notices are stored, there are only 6 pageid's (1-6), so you can see that if the page id is less than one, it displays an error message, if it is more than 6 it displays an error message.

This is all dandy for legal.php, but what about a database that is always getting new data, more and more pageid's, it would be a pain in the neck to keep changing the IF statement all the time, and would soon be confusing. What I am trying to accomplish is this: I want the page to check the database to see if the pageid that is in the url (ex: http://domain.com/page.php?pageid=21) exists in the database. If it does not, it displays the error, if it does contain the pageid, then it continues on with the rest of the code. I hope I am not being to confusing.

[shu]

//Try testing number of rows returned by your query
if (mysql_num_rows($display_title) > 0) {
  //Page exists...
} else {
  //No such page...
}

[Sharif]

Thanks for trying Shu but that doesn't work. It just shows a blank page for EVERYTHING (all pageid's).

[stolzyboy]

if ($pageid >6 || !$pageid)

[tomhath]

What Shu posted should work. If you're getting blank pages the script has some other error (probably didn't compile).

[stolzyboy]

i agree with tomhath, except for the compile part, since PHP is an interpreted language and not a compiled language like JAVA, but at any rate, if you count the rows with mysql_num_rows and it comes up with zero, the conditions should work, shu is correct, i don't think you are implementing it correctly, since this method has worked for many people to check to see if any records are returned

[Frederick]

the only thing i see wrong with your code is this line

if ($pageid < 1 OR $pageid >6)

use

if ($pageid < "1" || $pageid > "6")

or

if ($pageid < '1' || $pageid > '6')

|| works better then OR and the double and single quotes will insure no confilict with the if statement when you have more then one variable in the statment.

[stolzyboy]

all the double and single quotes are doing is testing a string or not, no need when the condition is checking an integer

[Sharif]

ok, can someone include Shu's code with my code, I just want to see how it should be put together because I can't figure it out.

[Frederick]

<?php
if ($pageid < 1 || $pageid > 6) {
echo "No Such Page";
} else {
&nbsp&nbsp$sql = mysql_query("SELECT * FROM legal WHERE pageid='$pageid' ")
&nbsp&nbspor die (mysql_error());
&nbsp&nbspwhile ($row = mysql_fetch_array($sql)) {
&nbsp&nbspecho "".$row["pagetitle"]."<br>";
&nbsp&nbsp}
}
?>

[tomhath]

This is all dandy for legal.php, but what about a database that is always getting new data, more and more pageid's, it would be a pain in the neck to keep changing the IF statement all the time, and would soon be confusing.

His original question was how to get rid of the hardcoded IF condition; he already has it working that way.

Sharif - post the code that always returned a blank page.

[Sharif]

tomhath,

here is my code mixed with Shu's code, this returns a blank page:

<?php
$display_title = mysql_query("SELECT pagetitle FROM legal WHERE pageid=$pageid")
or die ("MySQL Error - Please report this to the webmaster");
if (mysql_num_rows($display_title) > 0)
{
echo("No Such Page");
}
else
{
while ($row = mysql_fetch_array($display_title))
{
$legal = $row["pagetitle"];
echo $legal;
}
}
?>

What this is supposed to do is check the table to see if the pageid in the url exists in the table, if it doesn't it displays "No Such Page", it it does, it continues on with the rest of the code (inside the ELSE statement).

[shu]

The test should be the other way around:

$display_title = mysql_query("SELECT pagetitle FROM legal WHERE pageid=$pageid")
or die ("MySQL Error - Please report this to the webmaster");
if (mysql_num_rows($display_title) > 0) {
  while ($row = mysql_fetch_array($display_title)) {
    $legal = $row["pagetitle"];
    echo $legal;
  };

}  else {
  //No Such Page
}

[Sharif]

Hey Shu,

That works perfectly, and I don't even have to specify a limit in the pageid. Since I just started with PHP a week ago, I think it would be good to know what your code means, can you please explain?:

(mysql_num_rows($display_title) > 0)

Thanks for all the help.

[shu]

/*
If the query returns a record, that's if there exists a record in your database 
with the page_id = $pageid, it will return that 
record! Therefore the number of rows 
returned would be one!  Let me put it this way, maybe it's easy to understand:
*/

$num_rows = mysql_num_rows($display_title);

if ($num_rows > 0)  {
   //Yipee, found a page
} else {
  //No such page
}

[Sharif]

I sort of got that - see English for 14 yr olds are different 😃. Also, the "0" in if ($num_rows > 0), means FALSE?

[stolzyboy]

no it means that 0 (zero) results were returned