call another table from database

nrsh_ram · February 2, 2009

hi...good morning,

in my table, have "name, id " and "major"

name and id is from same table from database, major is from another table..

$cur_login = $_SESSION["username"];
$conn= db_connect();

$aa=0;	

$query = "select  mentee, username_mentee from mentormentee where username_mentor = '$cur_login'";
$result = $conn->query($query);

<?php
		while ($row = mysqli_fetch_assoc($result)){ 
		$aa = $aa + 1;
		?>	
		        <td><?php echo $aa; ?> </td>
        <td><?php echo $row['mentee']; ?> </td>
        <td><?php echo $row['username_mentee']; ?> </td>
        <td> </td>
      </tr><?php } ?>

its show data for name and id, for major how to call it for specific name and id?

Maq · February 2, 2009

its show data for name and id, for major how to call it for specific name and id?

What common fields do they both have? You're going to have to LEFT JOIN on those fields...

Have you tried any queries yet? The one you have looks like it has nothing to do with your question...

nrsh_ram · February 2, 2009

i have tried like this...but no error, for major its not showing the output.

$cur_login = $_SESSION["username"];
$conn= db_connect();

$aa=0;	

$query = "select  mentee, username_mentee from mentormentee where username_mentor = '$cur_login'";
$result = $conn->query($query);

$row1 = $username_mentee;
$query1 = "select major from mentee_personal_details where mentee_id = '$row1'";

<?php
		while ($row = mysqli_fetch_assoc($result)){ 
		$aa = $aa + 1;
		?>	
		        <td><?php echo $aa; ?> </td>
        <td><?php echo $row['mentee']; ?> </td>
        <td><?php echo $row['username_mentee']; ?> </td>
        <td><?php echo $row1['major']; ?> </td>
      </tr><?php } ?>

Maq · February 2, 2009

You need to replace common_field with whatever is common both tables to eliminate duplicates. If you don't know, how please dump both table structures and I will show you. Your query should look something like:

$query = "SELECT mm.mentee, mm.username_mentee, mp.major ".
"FROM mentormentee mm LEFT JOIN mentee_personal mp ".
"ON mm.common_field = mp.common_field".
        "WHERE mm.username_mentor = '$cur_login' AND mp.mentee_id = '$row1'";

*** If username_mentor AND mentee_id are both unique to the same thing then you only need to use one of them, which is probably the case.

nrsh_ram · February 2, 2009

table name mentor_personal_details

mentee_id | major |

012000000 | engineering |

012000001 | science |

table name mentormentee

username_mentor|mentee |username_mentee |

hod2 |mentee2 |012000000 |

hod2 |mentee2 |012000001 |

temp1 |mentee5 |012000003 |

Maq · February 2, 2009

Okay, I think you need to do

ON mp.menteeid = mm.username_menteee

nrsh_ram · February 2, 2009

<?php
while ($row = mysqli_fetch_assoc($result)){ 
		$aa = $aa + 1;
		?>

its show error for this line..i have to change it?

Maq · February 2, 2009

What's the error say? Post all of the code as well.

If the error is not specific please put this as your first PHP code:

ini_set ("display_errors", "1");
error_reporting(E_ALL);

nrsh_ram · February 3, 2009

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in C:\Program Files\xampp\htdocs\nrsh\all.php on line 154

nrsh_ram · February 3, 2009

$cur_login = $_SESSION["username"];
$conn= db_connect();

$aa=0;

$query1 = "SELECT mm.mentee, mm.username_mentee, mp.major ".
"FROM mentormentee mm LEFT JOIN mentee_personal_details mp ".
   "ON mp.mentee_id = mm.username_mentee".
        "WHERE mm.username_mentor = '$cur_login' AND mp.mentee_id = mm.username_mentee";	


$result = $conn->query($query1);

<?php
		while ($row = mysqli_fetch_assoc($result)){ 
		$aa = $aa + 1;
		?>	
        <td><?php echo $aa; ?> </td>
        <td><?php echo $row['mentee']; ?> </td>
        <td><?php echo $row['username_mentee']; ?> </td>
        <td><?php echo $row['major']; ?> </td>
      </tr><?php } ?>

the error is:

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in C:\Program Files\xampp\htdocs\nrsh\all.php on line 153

killah · February 3, 2009

Is both the code on the same .php page? and in the same function?

If not then it will not work.

nrsh_ram · February 3, 2009

yes..both is on same page.

Maq · February 3, 2009

$result = $conn->query($query1);

$result is not the correct type to pass to your mysqli_fetch_array(). Can you post your db_connect class?

nrsh_ram · February 3, 2009

db_connect?

whic one ?

<?php

function db_connect()
{
   $result = new mysqli('localhost', 'root', '', 'rama1'); 
   if (!$result)
     throw new Exception('Could not connect to database server');
   else
     return $result;
}

?>

Maq · February 3, 2009

Sorry didn't mean to say class, even though it should be.

What does the query function look like?

You're calling:

$result = $conn->query($query1);

which doesn't make sense. Can you show us more of your code? Preferably all of it if it's not too long (

nrsh_ram · February 3, 2009

<?php
require_once('include/session.php');
require_once('include/user_auth_fns.php');

$cur_login = $_SESSION["username"];
$conn= db_connect();

$aa=0;

$query1 = "SELECT mm.mentee, mm.username_mentee, mp.major ".
"FROM mentormentee mm LEFT JOIN mentee_personal_details mp ".
   "ON mp.mentee_id = mm.username_mentee".
        "WHERE mm.username_mentor = '$cur_login' AND mp.mentee_id = mm.username_mentee";	


$result = $conn->query($query1);

?>

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<link href= "2col_leftNav1.css" rel="stylesheet" type="text/css" />
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<link href="dropdowntabfiles/ddcolortabs.css" rel="stylesheet" type="text/css" />
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</head>

<body>

<div id="masthead">
  
   <div align="left"></div>
</div>

  <div id="masthead_1">
  <table width="300" border="0" align="right">
  <tr>
    <td><font class=label face="Verdana" size="2"><?php echo $_SESSION["name"]; ?> | time | Date |<a href="nlogout.php"> Logout</a>  </font></td>
  </tr>
</table>
<div id="colortab_1">
<div id="colortab" class="ddcolortabs">
<ul>
<li><a href="nprofile1.php" title="profile" ><span>PROFILE</span></a></li>
<li><a href="nforms.php" title="froms" ><span>FORMS</span></a></li>
<li><a href="view_report_menu.php" title="mentor" ><span>MENTOR</span></a></li>
<li><a href="nmanagementee.php" title="mentee" ><span>MENTEE</span></a></li>
<li><a href="mentor_setting.php" title="mysetting" ><span>MY SETTING</span></a></li>
</ul>
</div>
</div>
<div class="ddcolortabsline"> </div>                                   

  
  </div>
<div id="content"> 
  <div class="story">this features for own mentee 
    

-> all </div>
  <div id="breadCrumb_1">
    
    <table width="567" border="0" align="center">
      <tr>
        <td> </td>
        <td> </td>
        <td> </td>
        <td> </td>
      </tr>
      <tr>
        <td width="109" bgcolor="#FFD9D9"><div align="center">No</div></td>
        <td width="111" bgcolor="#FFD9D9">Name</td>
        <td bgcolor="#FFD9D9" width="149"><div align="center">Id</div></td>
        <td bgcolor="#FFD9D9" width="182"><div align="center">Major</div></td>
      </tr>
      <tr>
  <?php
		while ($row = mysqli_fetch_assoc($result)){ 
		$aa = $aa + 1;
		?>	
        <td><?php echo $aa; ?> </td>
        <td><?php echo $row['mentee']; ?> </td>
        <td><?php echo $row['username_mentee']; ?> </td>
        <td><?php echo $row['major']; ?> </td>
      </tr><?php } ?>
    </table>
  </div>
</div>

<div id="navBar">
  <div id="search"></div>

  <div class="relatedLinks">
    <h3> <div class="glossymenu">	
  <a href="nmanagementee_assign.php"  class="menuitem">Manage Mentee</a>
 <a class="menuitem submenuheader" href="#" >Own Mentee</a>	 
	<div class="submenu">
<ul>
<li><a href="#">All</a></li>
<li><a href="#">Name</a></li>
<li><a href="#">Id</a></li>
<li><a href="#">Cohort</a></li>
<li><a href="#">Course</a></li>	

</ul>

</div> 
<a href="#"  class="menuitem">Status</a>


</div> </h3>
  </div>
  </div>
</body>
</html>

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call another table from database

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