How to use like in this query

Question

How can i use LIKE in this query?

I have a problem with the syntax.

$query="SELECT id,dediche FROM table_dediche WHERE dediche not in (select dedica as dediche from ".$row['nome']."_quotes)";

I have to eliminate sentences from the table using LIKE %%.

        $query="SELECT id,dediche FROM table_dediche WHERE dediche not in (select dedica as dediche from ".$row['nome']."_quotes)";
        $sql = mysql_query($query); //faccio la query

        echo "<select style='width: 515px; margin-bottom:7px;' id='frasi' name='frasi'>";
        while ($res = mysql_fetch_array($sql)){ //prendo il risultato della query
        echo "<option value=\"".$res['dediche']."\">".$res['dediche']."</option>";
        }
        echo "</select>";
    ?>

Thanks in advance.

where is your sentence ? – Satish Sharma Jan 30 '14 at 10:23 — Satish Sharma, Jan 30 '14 at 10:23
where you want to apply like %% ? – Satish Sharma Jan 30 '14 at 10:28 — Satish Sharma, Jan 30 '14 at 10:28
I have to eliminate the select phrases in it. – Reverter Jan 30 '14 at 10:42 — Reverter, Jan 30 '14 at 10:42

Ryan · Answer 1 · 2014-01-30T10:33:57.953

1

To eliminate sentences, I presume you are looking for NOT LIKE

For example

SELECT sentence FROM sentences WHERE sentence NOT LIKE '%once upon a time%';

That will obviously need to be changed to work with your table structure.

edited Jan 30 '14 at 10:33

answered Jan 30 '14 at 10:17

Ryan

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@SatishSharma please refrain from editing my answer to add irrelevant content. – Ryan Jan 30 '14 at 10:34

score 0 · Answer 2 · edited May 23 '17 at 12:03

If I understand you correctly you have one table with words (or more general text). This is your table $nome. Now you want to select all rows from another table table_dediche, where column dediche does not containt anything that is in the table $nome. Is this correct?

That is not possible, if you get the values from another table. There is nothing like combining like and in with the result of a subquery. You can have a look at MySQL LIKE IN()? where they solve that problem using an regex. Not sure if that works in your case.

How to use like in this query

2 Answers2