How to using LIKE operator which data saved in variable?

Question

I need to search data in the database table.

I save data in a variable, then I want to display the data using it LIKE an operator when the data is called.

This is my code :

 $data_input_user = $inFilter;
 $data_input_user = mysql_real_escape_string($inFilter);
 $sqlGetFreq = "SELECT * FROM termfrequency WHERE keyword LIKE 
 CONCAT("%",$inFilter,"%") ";
 $freqResult = mysql_query($sqlGetFreq) or die(mysql_error());

Beside that i was trying this sql query

$sqlGetFreq = "SELECT keyword, frequency FROM termfrequency WHERE keyword LIKE '%$inFilter%' ";
$freqResult = mysql_query($sqlGetFreq) or die(mysql_error());

I was trying that code, but it failed.

Any idea please?

Thanks.

get rid of those old mysql_ functions, use prepared statements with mysqli. Then search SO for posts that talk about "prepared statements" and "LIKE" you WILL find a pre-existing solution on SO. — mickmackusa, Jun 19 '17 at 03:46
Possible duplicate of [Correct way to use LIKE '%{$var}%' with prepared statements? \[mysqli\]](https://stackoverflow.com/questions/28385145/correct-way-to-use-like-var-with-prepared-statements-mysqli) — mickmackusa, Jun 19 '17 at 03:55
I have tried this syntax $variabel_like = "%" . $inFilter . "%"; //$data_inputan_user = mysql_real_escape_string($inFilter); $sqlGetFreq = $variabel_like->prepare("SELECT * FROM termfrequency WHERE kata_kunci LIKE $inFilter"); $sqlGetFreq->bind_param("s",$inFilter); $sqlGetFreq->execute(); but i get error : Fatal error: Call to a member function prepare() on string in C:\xampp\htdocs\myfolder\phpfile.php — zordon_tri, Jun 19 '17 at 04:13
that answer from Correct way to use LIKE '%{$var}%' with prepared statements? [mysqli] doesn't work in my code — zordon_tri, Jun 19 '17 at 04:14
You must properly connect mysqli and get rid of ALL mysql_ functions. replace them with mysqli_ functions — mickmackusa, Jun 19 '17 at 04:14
You need to delete this duplicate question, spend some time researching / educating yourself about how to use mysqli and then come back and ask your question after trying to self-solve. The answer is already on SO. — mickmackusa, Jun 19 '17 at 04:15
Your accepted solution is bad practice, as I've been trying to tell you the whole time. I'm trying to help you find the "right way" to do this. — mickmackusa, Jun 19 '17 at 21:17
problem solved use this query : problem solved, i use this query : $sqlGetFreq = "SELECT * FROM termfrequency WHERE keyword IN ($inFilter) ORDER BY keyword,tittle_symbol"; $freqResult = mysql_query("SELECT * FROM termfrequency WHERE keyword IN ($inFilter) ORDER BY keyword,titte_symbol ASC") or die(mysql_error()); — zordon_tri, Sep 05 '17 at 07:12

score -2 · Accepted Answer · answered Jun 19 '17 at 06:33

-2

Try this

   $sqlGetFreq = "SELECT keyword, frequency FROM termfrequency WHERE keyword LIKE '%".$inFilter."%' ";

Also please set the following to track any possible error.

ini_set("display_errors", "1");
error_reporting(E_ALL);

answered Jun 19 '17 at 06:33

Sehdev

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ok, now nothing error, but when all mysql I replace mysqli data not display – zordon_tri Jun 19 '17 at 08:32
mysqli_query() takes connection variable as first parameter. You might be missing that. mysqli_query($con,$sql) – Sehdev Jun 19 '17 at 09:07
it solved query : problem solved, i use this query : $sqlGetFreq = "SELECT * FROM termfrequency WHERE keyword IN ($inFilter) ORDER BY keyword,tittle_symbol"; $freqResult = mysql_query("SELECT * FROM termfrequency WHERE keyword IN ($inFilter) ORDER BY keyword,titte_symbol ASC") or die(mysql_error()); – zordon_tri Sep 05 '17 at 07:12

How to using LIKE operator which data saved in variable?

1 Answers1