[yuva17raj]

hi
i am trying to update my sql table data with the csv/excel file data
my sql table is as follows

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1. test test1
2. 23   usd
3. 25   ero
4. 34   test
5.  

and my csv file has

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1. 34 testing
2. 23 usding
3.  

so here i have to update 34 and 23 as in the csv file
i am trying with the code as follows

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1. $link = mysql_connect('localhost', 'root', 'root') or die('Could not
2. connect: ' . mysql_error());
3. $handle = fopen("C:/testexcel.csv","r");
4.  
5. // loop content of csv file, using comma as delemiter
6. while (($data = fgetcsv($handle, 4096, ",")) !== FALSE)
7.  {
8.     $test= $data[0];
9.     $test1= $data[1];
10. //$price2 = floatval($data[2]);
11.  
12.     $query = "SELECT * FROM testdb";
13. //$test1 = mysql_num_rows($query);
14. //echo $test1;
15.  
16.         if (!$result = mysql_query($query))
17.          {
18.             continue;
19.             }
20.  
21.     if ($line = mysql_fetch_array($result, MYSQL_ASSOC))
22.      {
23. // entry exists update
24. //$query = "UPDATE testdb SET test='$dist' WHERE test=$test1";
25.     mysql_query("UPDATE testdb SET test = '$test1' WHERE phoneno= '$phone'");
26.     if (mysql_affected_rows() <= 0)
27.      {
28.         echo "testing sucessfull";
29.         }
30.     } 
31.     else {
32.         // entry don't exists continue or insert...
33.  
34.             }
35.  
36. mysql_free_result($result);
37. }
38.  
39. fclose($handle);
40. mysql_close($link);
41.  
42. //error_reporting(0) ;// show no warnings
43. error_reporting(E_ALL); //show all warnings.
44.  
45. ?>
46.  

and i have tried with the above code
so kindly help me with the solution as it is not updating to the database table
regards
yuvaraj

[code green]

[PHP]mysql_query("UPDATE testdb SET test = '$test1' WHERE phoneno= '$phone'");[/PHP] I don't see a $phone variable

[yuva17raj]

that was my actual code and i have posted my sample with this post kindly reply me with the solution



<?
$link = mysql_connect('localhost', 'root', 'root') or die('Could not
connect: ' . mysql_error());
$handle = fopen("C:/testexcel.csv","r");

// loop content of csv file, using comma as delemiter
while (($data = fgetcsv($handle, 4096, ",")) !== FALSE)
{
$test1= $data[0];
$test2 = $data[1];

$query = "SELECT someF2 FROM mytestdb";
if (!$result = mysql_query($query))
{
continue;
}

if ($line = mysql_fetch_array($result, MYSQL_ASSOC))
{
mysql_query("UPDATE mytestdb SET someF1 = '$test1' WHERE someF2= '$test2'");
if (mysql_affected_rows() <= 0)
{
echo "testing sucessfull";
}
}
else {
// entry don't exists continue or insert...
}

mysql_free_result($result);
}

fclose($handle);
mysql_close($link);

error_reporting(E_ALL); //show all warnings.

fclose($handle);

?>


this how i have tried but my database table is not updating so provide me a solution

regards
yuvaraj

[code green]

You do not throw a load of code at us and demand an answer.
To find a solution we need to know what is happening.
I have not checked your syntax, I have better things to do, but I suspect the form is not passing the data as you think.
Also, this need to be a the beginning of your script

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1. error_reporting(E_ALL); //show all warnings.

Not the end.
Please put some echo statements in there that displays the contents of your variables and arrays,
then maybe the problem will become obvious

[Dormilich]

you could try var_dump($query) to see if the MySQL syntax is correct (i.e. all the php variables are replaced (I have my doubts about '$test' because single quoted strings are not parsed))

regards

[yuva17raj]

hi
thanks for ur reply i got it with some other code in php
regards
yuvaraj