Searching a mysql database

[senyo]

Hi I am stupid and I can't search a database, everything I tried doesn't work, I want to search the table client, field name.

[RaythMistwalker]

Hi I am stupid and I can't search a database, everything I tried doesn't work, I want to search the table client, field name.

mysql_connect(HOST, USERNAME, PASSWORD);
mysql_select_db(DB NAME)

$qry = "SELECT name FROM client";
$result = mysql_query($qry);

[senyo]

I am using this but I am getting a blank screen

mysql_connect("localhost","2","2");

mysql_select_db("book");
$each="john";
$qry = "SELECT FROM client where client.name like '%".$each."%' ";
$result = mysql_query($qry);

echo $result;

[senyo]

I have this 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 'FROM client where name like john' at line 1

 

What does it mean?

[RaythMistwalker]

mysql_connect("localhost","2","2");

mysql_select_db("book");
$each="john";
$qry = "SELECT FROM client WHERE client.name LIKE '%".$each."%' ";
$result = mysql_query($qry);

echo $result;

 

WHERE and LIKE have to be in caps.

[Maq]

mysql_connect("localhost","2","2");

mysql_select_db("book");
$each="john";
$qry = "SELECT FROM client WHERE client.name LIKE '%".$each."%' ";
$result = mysql_query($qry);

echo $result;

 

WHERE and LIKE have to be in caps.

 

No they don't.  The problem is that you're not SELECTing anything.  Moving to MySQL Help section.

[Lamez]

SELECT * FROM ...

[senyo]

how to finish the code so that it returns data from the database?

[Maq]

how to finish the code so that it returns data from the database?

 

You need to provide the SELECT statement data to select.  Read more here:

 

http://dev.mysql.com/doc/refman/5.0/en/select.html

[senyo]

I have this and it shows no errors only this: Resource id #4Resource id #5

how to make it show the whole entry?

mysql_connect("localhost","2","2");

mysql_select_db("book");
$each="john";
$qry = "SELECT * FROM client WHERE client.name LIKE '%".$each."%'";
$result = mysql_query($qry);


$eror= mysql_query($qry) or die(mysql_error()); 
$sebuty=$result['name'];
echo $sebuty;
echo $result;

echo $eror;

[aebstract]

mysql_connect("localhost","2","2");

mysql_select_db("book");
$each="john";
$result= "SELECT * FROM client WHERE client.name LIKE '%".$each."%'";
while($row = mysql_fetch_array($result))
  {
  $sebuty= $row['name'];
  }
echo $sebuty;
echo $result;

echo $eror;

 

Need to fetch the information.

http://php.net/manual/en/function.mysql-fetch-array.php

 

[senyo]

now I get this mysql_fetch_array() expects parameter 1 to be resource, string given in line 12 that is

while($row = mysql_fetch_array($result))

[Maq]

$result= "SELECT * FROM client WHERE client.name LIKE '%".$each."%'";
while($row = mysql_fetch_array($result))

 

should be:

 

$sql= "SELECT * FROM client WHERE client.name LIKE '%".$each."%'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))

 

You should also add in some error handling.  Read this blog for more details:

http://www.phpfreaks.com/blog/or-die-must-die

[senyo]

Thanks it work

 

<?php

mysql_connect("localhost","2","2");

mysql_select_db("book");
$each="john";
$sql= "SELECT * FROM client WHERE client.name LIKE '%".$each."%'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
  {
  $fromdatabase= $row['name'];
  echo $fromdatabase;
  }


?>

[aebstract]

Oh, doh.. forgot that line 

[Maq]

Glad it's working.  Keep in mind, if you're using user input in your query you must invoke mysql_real_escape_string on the value to prevent mysql injections.