[tMike]

Hi there,

i've work on it a long time and it will not work (I use StrickenKids MySQL Plugin).

I tested these query

Код:

SELECT * FROM `help` WHERE `searchWord1` LIKE '%%s%' LIMIT 0 , 30

For example i type in "hous", it should look like this

Код:

SELECT * FROM `help` WHERE `searchWord1` LIKE '%hous%' LIMIT 0 , 30

I tested a lot of methods, for example this one:

Код:

SELECT * FROM `help` WHERE `searchWord1` LIKE '% %s %' LIMIT 0 , 30

This will printed in this form: "% hous %", but the mysql-database don't know what to do with it.

I have saved "house" in searchWord1, i want to make a search system on my server.
The database should also find the content when I use the name of the column specified imprecise.

When i type in "hous" as a search-word it should find the content with "house" (with the sql field in phpmyadmin it works perfectly, but not with mysql_query, - maybe it lay on the strings?).

I've printed it and this will come out:

Код:

SELECT * FROM `help` WHERE `searchWord1` LIKE '%s' LIMIT 0 , 30

(only the percent and the S, not the string which i have defined!)

Thanks for help, sorry for my bad english.

[Calgon]

Try using a double percentage sign. If you're using format(), you need to escape the percentage sign properly or it'll cause problems.

A single percentage sign in format() is considered as a placeholder, more than 1 will actually show in the string properly.

Код:

SELECT * FROM `help` WHERE `searchWord1` LIKE '%%%s%%' LIMIT 0 , 30

[tMike]

Thanks a lot, genius.