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what is the function to count all the fields in a table?

Posted on 2008-06-09
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Last Modified: 2010-04-21
lets say i have a table named t1
and it have in this table
id -  place
1 -   india
2 -  florida
3 -  la
4 -  new dom

so when i use the function it will return the nmber 4 for me

so what is this function?
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Question by:mgtm3
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Expert Comment

by:hernst42
Comment Utility
the SQL is
"SELECT count(*) FROM t1";
to count all rows in that table
mysql_connect("localhost", "mysql_user", "mysql_password") or

    die("Could not connect: " . mysql_error());

mysql_select_db("mydb");

$result = mysql_query("SELECT count(*) FROM t1");

$row = mysql_fetch_array($result, MYSQL_NUM);

echo $row[0];  

mysql_free_result($result);

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Author Comment

by:mgtm3
Comment Utility
i am trying to to what u told me
1-i am using mysqli to connect to the sever
2-
                     --this what it did and didnt work ---
$result = mysqli_query("SELECT count(*) FROM pollnames");
$row = mysqli_fetch_array($result, $link);
echo $row[0];  
-----------------------------------------------------------

i got this error


Warning: mysqli_query() expects at least 2 parameters, 1 given in C:\wamp\www\poll\222.php on line 29

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\wamp\www\poll\222.php on line
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LVL 48

Expert Comment

by:hernst42
Comment Utility
then its:

$result = mysqli_query($link, "SELECT count(*) FROM pollnames");
$row = mysqli_fetch_array($result, $link);
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Author Comment

by:mgtm3
Comment Utility
now i got this error
Warning: mysqli_fetch_array() expects parameter 2 to be long, object given in C:\wamp\www\poll\222.php on line 30
this is what i wrote
 

$result = mysqli_query($link, "SELECT count(*) FROM pollnames");

$row = mysqli_fetch_array($result, $link);

echo $row[0]; 

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Accepted Solution

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hernst42 earned 500 total points
Comment Utility
That was what I copied from your code. Not realy famliliar with mysqli usage. This way should now work:
$result = mysqli_query($link, "SELECT count(*) FROM pollnames");

$row = mysqli_fetch_array($result, MYSQLI_NUM);

echo $row[0]; 

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Author Closing Comment

by:mgtm3
Comment Utility
thanks
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