getting out 1 field.

[IwnfuM]

i was trying to getting out 1 field from the db.

i was doin it like this .

$getout = mysql_query("SELECT `ip` FROM `table` WHERE `time`='".$time."'");

it doesnt getting out what i asked for.

can any one help me with this.

thanks , Mor.

[PFMaBiSmAd]

Time is fairly precise. Are you sure you have a matching row where your time column has a value that exactly matches what is in your $time variable?

 

Generally when trying to match time, you use <, <=, >, or >= comparisons rather than just an = comparison.

[IwnfuM]

i double check it twice.

it gives me.

"Resource id #" with a number.

anyway , if that important i was just echoing the

$getout

.

thanks , Mor.

 

[PFMaBiSmAd]

mysql_query() returns a result resource. You must fetch the data from the result resource before you can use it. There is a family of mysql_fetch_xxxx() functions that will let you do that.

 

I recommend reading a basic mysql tutorial or read the examples in the php documentation to learn how to execute a query and fetch the data from the query.

[IwnfuM]

made it .

i got 1 more question.

what is the difference from Mysql_fetch_array to Mysql_fetch_assoc .

thanks , Mor.

[Pikachu2000]

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