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php to mysql

Posted on 2009-02-21
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Last Modified: 2012-05-06
in a .txt document

firstname

lastname

email address


firstname

lastname

email address


how could someone read this data in php and put it on a mysql table


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Question by:jerseyguy0
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15 Comments
 
LVL 143

Accepted Solution

by:
Guy Hengel [angelIII / a3] earned 200 total points
ID: 23700410
just to clarify: the data is 1 line at a time, or are the 3 information in 1 line each time? in the latter case, what is the separator?

now: EE is not a rent-a-coder, so you should try to clarify/check out the different parts of such a "simple" problem yourself:
* read a file
* split a string into parts (not necessarily required here...)
* write a sql statement for the INSERT
* run the sql statement

so, read up the sites php.net and mysql.com to get basic samples, and come back with concrete problems you have then, posting the code you have then, and the errors/problems you run into.
0
 
LVL 1

Author Comment

by:jerseyguy0
ID: 23700448
one line as a separator between firstname, lastname and email address

two lines as a separator between block of firstname lastname and email address
0
 
LVL 8

Expert Comment

by:agamal
ID: 23700455
please post a sample text file ... to send you the exact processing code :D
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LVL 3

Assisted Solution

by:OrTzabary
OrTzabary earned 200 total points
ID: 23700466
Read the file with fopen & fread (read in php.net)

after that explode the string with \n (new line seperator)

do foreach to work over the array and insert into the database with mysql query

example included...
$filename = "/usr/local/something.txt";
$handle = fopen($filename, "r");
$contents = fread($handle, filesize($filename));
fclose($handle);
 
$content_array = explode("\n", $contents);
 
foreach ($content_array AS $name => $value) {
mysql_query("INSERT INTO table SET field = '$value'");
}

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0
 
LVL 1

Author Comment

by:jerseyguy0
ID: 23700482
attached is txt file of firstname lastname emailadress
firstnamelastnameemail.txt
0
 
LVL 19

Assisted Solution

by:Michael701
Michael701 earned 400 total points
ID: 23700497
Something like this should get you started. I'll have to verify the field names of your database table.


$link = mysql_connect("localhost", "mysql_user", "mysql_password") 
       or die("Could not connect: " . mysql_error()); 
 
$handle = fopen ("/tmp/inputfile.txt", "r"); 
while (!feof ($handle)) { 
   $first_name = mysql_real_escape_string(fgets($handle)); 
   $last_name =  mysql_real_escape_string(fgets($handle)); 
   $email =      mysql_real_escape_string(fgets($handle)); 
   
   $sql_command="insert into customers (first_name, last_name, email ) values ('$first_name', '$last_name', '$email')";
   $ok=mysql_query($sql_command);
} 
fclose ($handle); 

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0
 
LVL 8

Assisted Solution

by:agamal
agamal earned 400 total points
ID: 23700589
try this ... change the variable of course

<?php
//reading file
$filename = "firstnamelastnameemail.txt";
$file = file($filename);
 
//database settings
$db = "test";
$table = "users";
$col1 = "first";
$col2 = "last";
$col3 = "mail";
 
//connect to mysql
$connection = mysql_connect('localhost', 'root', 'root') or die (mysql_error());
$db = mysql_select_db('test', $connection) or die (mysql_error());
 
//do the reading loop
$j = 0;
$result = array();
for ($i=0;$i< count($file);$i++) {
if ($j == 3)
{
//do the sql query
$query  = "insert into $table (`$col1` , `$col2` , `$col3` ) values ('$result[0]', '$result[1]', '$result[2]')";
$res = mysql_query($query) or die('Error, query `' . $query . '` failed');
$j = 0;
}
if ($file[$i] != "\n") {
$result[$j] = $file[$i];
$j++;
}
        }
 
        mysql_close($connection);
 
 
?>

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0
 
LVL 111

Assisted Solution

by:Ray Paseur
Ray Paseur earned 400 total points
ID: 23700674
I downloaded the test data.  Are there really blank lines in the input file?  Thanks, ~Ray
0
 
LVL 1

Author Comment

by:jerseyguy0
ID: 23700744
to ray_passeur

yes there are blank lines in the input file
1 blank line between firstname lastname emailaddress

2 blank lines between new people
0
 
LVL 1

Author Comment

by:jerseyguy0
ID: 23700746
to agamal:
try this ... change the variable of course


which variable should I change
0
 
LVL 39

Assisted Solution

by:Roger Baklund
Roger Baklund earned 400 total points
ID: 23700809
I would do something like this:
mysql_connect('localhost','user','pass') or die('Could not connect: '.mysql_error());
mysql_select_db('dbname') or die("Could not select database '$dbname'");
$data = file_get_contents('firstnamelastnameemail.txt');
$records = explode("\n\n\n",$data);
$inserted = 0;
foreach($records as $record) {
  list($fname,$lname,$email) = explode("\n\n",$record); 
  $fname = mysql_real_escape_string($fname);
  $lname = mysql_real_escape_string($lname);
  $email = mysql_real_escape_string($email);
  $res = mysql_query("insert into MyTable set fname='$fname',lname='$lname',email='$email'");
  if(!$res) echo 'Failed: '.$fname.' '.$lname.' '$email.'<br />'.mysql_error().'<br />';
  else $inserted++;
}
echo $inserted.' rows inserted.';

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0
 
LVL 39

Assisted Solution

by:Roger Baklund
Roger Baklund earned 400 total points
ID: 23700820
Sorry, error in line 12, a missing dot, it should be:

if(!$res) echo 'Failed: '.$fname.' '.$lname.' '.$email.'<br />'.mysql_error().'<br />';
0
 
LVL 111

Assisted Solution

by:Ray Paseur
Ray Paseur earned 400 total points
ID: 23700829
How many records do you have in the live data input file?
0
 
LVL 19

Assisted Solution

by:Michael701
Michael701 earned 400 total points
ID: 23700894
you could just add a few read lines to my code

$first_name = mysql_real_escape_string(fgets($handle)); 
$blank=fgets($handle);
$last_name =  mysql_real_escape_string(fgets($handle)); 
$blank=fgets($handle);
$email =      mysql_real_escape_string(fgets($handle)); 
$blank=fgets($handle);
$blank=fgets($handle);

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0
 
LVL 8

Assisted Solution

by:agamal
agamal earned 400 total points
ID: 23702266
you file name in
$filename = "firstnamelastnameemail.txt";
 
//database settings
Data base name
$db = "test";

table name
$table = "users";

first name column in your table
$col1 = "first";

last name column in your table
$col2 = "last";

email column in your table
$col3 = "mail";
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