I have a variable called $uid
, and I would like to print a message if the variable $uid is not contained in the "loginid" field of a MySQL table "tweets" on any row that has another variable, $submissionid
. I think my query $tweetquery
is okay. Basically, I would like the message to echo if the loginid variables pulled from "tweets" never equal $uid.
How can I do this?
Thanks in advance,
John
$tweet开发者_Go百科query = "SELECT loginid
FROM tweets
WHERE submissionid = '$submissionid'";
$tweetresult = mysql_query($tweetquery);
if...
echo '<div>Message</div>';
Have a look at mysql_num_rows($result)
.
if ( mysql_num_rows($result) > 0 )
{
// Do something
}
else
{
echo 'No results';
}
Call me crazy, but using mysql_num_rows
is not the way I'd do this. mysql_fetch_array
does not do anything destructive if nothing is returned, so call that as normal.
$row = mysql_fetch_array( $resource );
if(!$row)
{
//do what you would do if the query returned nothing
}
else
{
do
{
// do what you would do with the row
} while( $row = mysql_fetch_array( $resource ) );
}
The problem with mysql_num_rows
is that it actually represents a call to the database asking for how many rows exist in the pointer. It is a, frankly, needless back-and-forth which can easily be avoided. So why not avoid it?
You can use mysql_num_rows
:
if (mysql_num_rows($tweetresult) > 0)
// results found
Use mysql_num_rows()
:
if (mysql_num_rows($tweetresult) == 0) {
echo 'Message';
}
if(mysql_num_rows($tweetresult) == 0)
echo "<div>Message</div>";
else {
//Do something if $uid was found
}
mysql_num_rows()
returns the numbers of rows fetched from the database. If rows fetched are zero echo
a message.
$tweetquery = "SELECT loginid
FROM tweets
WHERE submissionid = '$submissionid'";
$tweetresult = mysql_query($tweetquery);
if(mysql_num_rows($tweetresult) > 0){
// your results here
} else {
echo '<div>Message</div>';
}
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