Search data between two specific year?

Hi All,

I have tabel called "candidate" in that i have two fields call Graduation Passing year and Post Graduation Passing year.
In that table i have more than thousands number of candidate. Here i want to make a search php form. Where i can get data of specific candidate who pass between specific year.

I have make a form for search with below fields:
Graduation Candidate search : Year From and Year To fields and submit button to call query form and display result.

however i tried my end but not able to make such query into Mysql. Please help me make mysql query to display candidate as per my year criteria search.

Jimi
---------------
Here is first search.php code
// search Graduation result

Graduation candidate seach<BR>
<form name="search1" action="graduation.php" method=post>
From : <input type=text size="10" name=from value=""> <BR><BR>
To : <Input type=text size="10" name=to value=""><BR><BR>
<center><input type="submit" value="Submit"></center>
</form>
---------------
2) gaduation.php

<?php

$loginname = "";

$password = "";

$database = "";

$con = mysql_connect("localhost", "$loginname", "$password") or die("could not connect");
mysql_select_db("$database");

$query = "SELECT * FROM candidate_foreign WHERE value BETWEEN" .$_post['from']. "AND '".$_post['to']."'";

$result = mysql_query ("$query");

echo"<table border=1><TH>Year</TH>";

while ($in = mysql_fetch_array($result))
{
echo"<tr>";
echo"<td>". $in['value'] . "<td>";
echo"</tr>";
}

echo "</table>";

mysql_close($con);
?>
---------------------
Getting error as - Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/graduation.php on line 18

Please help me to get appropriate result from the query.

Jimi

 

$query = "SELECT * FROM candidate_foreign WHERE `Graduation Passing year` BETWEEN" .$_post['from']. "AND '".$_post['to']."'";

$query = "SELECT * FROM candidate_foreign WHERE `Post Graduation Passing year` BETWEEN" .$_post['from']. "AND '".$_post['to']."'";

 

thks for your reply but getting an error as below:

mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/graduation.php on line 18

What i do now? and also i want to check data from "value" column...

JIMI

 

$query = "SELECT * FROM candidate_foreign WHERE value BETWEEN<space>" .$_post['from']. "<space> AND<space> '".$_post['to']."'";

 

now i have following two pages and running without getting any error. But not giving me any result except column name as "year". Please check code of two php file below:
-----------------------------------------
1) search.php

// search Graduation result

Graduation candidate seach<BR>
<form name="search1" action="graduation.php" method=post>
From : <input type=text size="10" name=from value=""> <BR><BR>
To : <Input type=text size="10" name=to value=""><BR><BR>
<center><input type="submit" value="Submit"></center>
</form>
-----------------------------------------
2) graduation.php
<?php

$loginname = "";

$password = "";

$database = "";

$con = mysql_connect("localhost", "$loginname", "$password") or die("could not connect");
mysql_select_db("$database");

$query = "SELECT * FROM candidate_foreign WHERE value BETWEEN '" .$_post['from']. "' AND '".$_post['to']."'";

$result = mysql_query ("$query");

echo"<table border=1><TH>Year</TH>";

if(mysql_num_rows($result)){
while ($in = mysql_fetch_array($result))
{
echo"<tr>";
echo"<td>". $in['value'] . "<td>";
echo"</tr>";
}
}

echo "</table>";

mysql_close($con);

?>

 

I think code is ok just prb with query
Try this
$query = "SELECT * FROM candidate_foreign WHERE value BETWEEN " .$_post['from']. " AND ".$_post['to'];

 

now getting an error like as below:
mysql_num_rows(): supplied argument is not a valid MySQL result resource in /var/www/graduation.php on line 18

May be now query is wrong. What u say?

JIMI

 

try to check $_post['from'] and $_post['to'] before use them:

echo 'from='.$_post['from'];
echo 'to='.$_post['to'];

what's the result?

 

write

echo $query ; check what are getting then try to run this line in mysql(phpmyadmin or .....)

 

I got blank result. Is there any wrong in my code.

Jimi

 

I got blank result. Is there any wrong in my code?

Jimi

 

place the query in mysql and check what its results ....

 

Without value reference from search form i run that graduation.php i got result . what would be the problem in my code?
I run below query
---------------
<?php

$loginname = "";

$password = "";

$database = "";

$con = mysql_connect("localhost", "$loginname", "$password") or die("could not connect");
mysql_select_db("$database");

$query = "SELECT * FROM candidate_foreign WHERE value BETWEEN 2000 and 2006;

$result = mysql_query ("$query");

echo"<table border=1><TH>Year</TH>";

if(mysql_num_rows($result)){
while ($in = mysql_fetch_array($result))
{
echo"<tr>";
echo"<td>". $in['value'] . "<td>";
echo"</tr>";
}
}

echo "</table>";

mysql_close($con);

?>
--------------
Jimi
Just want to assign value of 2000 and 2006 from my search form.

 

Print these lines at front of graduation.php

echo "<pre>";
print_r($_POST);

and check that are you getting ur post values ?

 

$_post['from']. " AND ".$_post['to'];

Capital
$_POST['from']
$_POST['to']

 

thks its work and giving me result now...

thank you your urgent help

Hai friend i have one more query regarding php. Would u like to help me in that.

I have code with that i am paging my data. I just want to sort my data as per every column title. Do u have any idea how would i sort data through column name. I have mulitple column in my table.

Hope you would help in this prob too..

Regards,
JIMI

 

yes i think it is the easiest way to check where the problem is in the PHP code or in SQL query try this ...

 

I just inner join my table with other table and run below query now again its not giving me year. What wrong in query below:


"SELECT `first_name`, `last_name`, `address1`, `address2`, `city`, `state`, `current_employer`, `entered_by`, `owner`, `email1` FROM candidate INNER JOIN candidate_foreign ON candidate.candidate_id = candidate_foreign.assoc_id WHERE candidate_foreign.field_name = 'IITYear' AND value >= '" .$_POST['from']. "' AND value <= '".$_POST['to']."'";

 

" .$_POST['from']. " AND value <= ".$_POST['to']


Remove quates form $_POST['from'] and $_POST['to']

 

Once again i am here for need ur help...

I tried mentioned code below at my end:

--------
<?php

echo 'from='.$_POST['from'];
echo 'to='.$_POST['to'];

$loginname = "";

$password = "";

$database = "";

$con = mysql_connect("localhost", "$loginname", "$password") or die("could not connect");
mysql_select_db("$database");

$query = "SELECT 'first_name', 'last_name', 'address1', 'address2', 'city', 'state', 'current_employer', 'entered_by', 'owner', 'email1' FROM candidate INNER JOIN candidate_foreign ON candidate.candidate_id = candidate_foreign.assoc_id WHERE candidate_foreign.field_name = 'IITYear' AND value >= '" .$_POST['from']. "' AND '".$_POST['to']."'";

$result = mysql_query ("$query");

echo"<table border=1><TH>First Name</TH><TH>Address</TH><TH>Last Name</TH><TH>City</TH><TH>State</TH><TH>Email</TH>";

while ($in = mysql_fetch_array($result))
{
echo"<tr>";
echo"<td>". $in['first_name'] . "<td>";
echo"<td>". $in['address1'] . "<td>";
echo"<td>". $in['last_name'] . "<td>";
echo"</tr>";
}


echo "</table>";

mysql_close($con);

?>
-------
When i run at my end with help of search.php . I have been not getting any error. but also i not getting any value from table too. I am getting error as below:
----------
from=2006to=2009

First Name Address Last Name City State Email
first_name address1 last_name
first_name address1 last_name
first_name address1 last_name
first_name address1 last_name
first_name address1 last_name
first_name address1 last_name
first_name address1 last_name
first_name address1 last_name
first_name address1 last_name
first_name address1 last_name
--------------

Please help as u did last day.

JIMI