Inserting data into table_name as php variable

Question

I am taking table name as php variable to insert data into table. But it gives error. What's bad here?

            if($flag == 1)
                $table = 'frrole_pupolar_article';
            else
                $table = 'frrole_category_article';
            $insertQuery1 = "INSERT INTO '.$table.' (`url`, `sentiment`, `category`, `title` ,`time`,`img_url`,`rt_count`,`tweet_count`) VALUES ('".$url."','".$setiment."','".$category."','".$title."','".$time."','".$img_url."','".$rt_count."','".$tweet_count."')";

error:

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''.frrole_category_article.' (`url`, `sentiment`, `category`, `title` ,`time`,`im' at line 1

if all your columns and values are the same why are you putting this in different tables? just add a 'type' column and make your $table = type. I don't see why you need to have same data types in diff tables. Re think your approach. — Jakub
– Jakub, Commented Jan 13, 2014 at 14:23

sas · Accepted Answer · 2014-01-13 05:48:58Z

1

$insertQuery1 = "INSERT INTO '" .$table. "' (`url`, `sentiment`, `category`, `title` ,`time`,`img_url`,`rt_count`,`tweet_count`) VALUES ('".$url."','".$setiment."','".$category."','".$title."','".$time."','".$img_url."','".$rt_count."','".$tweet_count."')";

You wrongly written '.$table.' instead of '" .$table. "'

answered Jan 13, 2014 at 5:48

sas

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2 Comments

user123 Over a year ago

thanks but still gives error 'Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''frrole_category_article' (url, sentiment, category, title , time , `i' at line 1'

sas Over a year ago

$insertQuery1 = "INSERT INTO '" .$table. "' (url, sentiment, category, title, time, img_url, rt_count, tweet_count) VALUES ('".$url."','".$setiment."','".$category."','".$title."','".$time."','".$img_url."','".$rt_count."','".$tweet_count."')";

try this

zzlalani · Accepted Answer · 2014-01-13 05:53:13Z

1

Just do

$insertQuery1 = "INSERT INTO $table (`url`, `sentiment`, `category`, 
                `title` ,`time`,`img_url`,`rt_count`,`tweet_count`) 
                 VALUES ('".$url."','".$setiment."','".$category."',
                 '".$title."','".$time."','".$img_url."','".$rt_count."',
                 '".$tweet_count."')";

remove the concatenations and single quotes

answered Jan 13, 2014 at 5:53

zzlalani

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2 Comments

user123 Over a year ago

it says : Error: Table 'mysql.frrole_category_article' doesn't exist

zzlalani Over a year ago

remove mysql. out of your table name frrole_category_article

Jakub · Accepted Answer · 2014-01-13 05:48:42Z

0

If your string is $insertQuery1 = "INSERT INTO '.$table.' (url,sentiment, You cannot escape it by using single quotes '..'

Just do double quotes:

$insertQuery1 = "INSERT INTO ".$table." (`url`, `sentiment`, ....

also for all that is holy, use damn {} for if(){} statements, not using them is poor form

answered Jan 13, 2014 at 5:48

Jakub

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2 Comments

Jakub Over a year ago

Not if you have to read the code afterwards and everything is a mess.

user123 Over a year ago

it says Error: Table 'mysql.frrole_category_article' doesn't exist though table exist

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