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mtvaran

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Posts posted by mtvaran

    how to fix "where clause is ambiguous" error!!

    in PHP Coding Help

    Posted

    this is my full query for searching data from multiple table. still i could not get any result. could anyone check please where am i made mistake?

    <?php  

	$con = mysql_connect("localhost","root","");
if (!$con)
  	{
  	die('Could not connect: ' . mysql_error());
  	}

mysql_select_db("uni", $con);

$sql = "
    SELECT take.StudentID
          ,student.StudentName
          ,take.CourseID
          ,course.CourseName
    FROM take
        ,student
        ,course
    WHERE take.StudentID = student.StudentID
    AND take.CourseID = course.CourseID
    AND take.StudentID LIKE '$_POST[sid]%'
    ORDER BY take.StudentID ASC
";

$result = mysql_query($sql) or trigger_error('MySQL Error: ' . mysql_error(), E_USER_ERROR);

echo"<br>";

echo "<center><table width=700 border=1>";
echo "<tr><th>StudentID</th><th>StudentName</th><th>CourseID</th><th>CourseName</th></tr>";

while($row = mysql_fetch_array ($result))
echo mysql_error();
{

echo "<tr><td>";
echo $row['StudentID'];
echo "</td><td>";
echo $row['StudentName'];
echo "</td><td>";
echo $row['CourseID'];
echo "</td><td>";
echo $row['CourseName'];
echo "</td></tr>";

}

echo "</table>";

mysql_close($con); 
?>

     

     

    how to fix "where clause is ambiguous" error!!

    in PHP Coding Help

    Posted

    hi guys how to fix this error. i have this query

    $sql = "
    SELECT take.StudentID
          ,student.StudentName
          ,take.CourseID
          ,course.CourseName
    FROM take
        ,student
        ,course
    WHERE take.StudentID = student.StudentID
    AND take.CourseID = course.CourseID
    AND StudentID LIKE '$_POST[sid]%'
    ORDER BY StudentID ASC
";

$result = mysql_query($sql) or trigger_error('MySQL Error: ' . mysql_error(), E_USER_ERROR);

     

     

     

    error msg: Fatal error: MySQL Error: Column 'StudentID' in where clause is ambiguous in C:\wamp\www\dis_take.php on line 98

     

    searching data column wise!

    in Applications

    Posted

    could anyone please check this code for me. i get the error msg like...

    Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\dis_study.php on line

     

     

    <?php 

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }


mysql_select_db("uni", $con);

$result = mysql_query("SELECT study.StudentID,student.StudentName,study.ProgrammeName FROM study,student WHERE study.StudentID = student.StudentID AND StudentID LIKE '$_POST[sid]%' ORDER BY StudentID ASC");

echo"<br>";

echo "<center><table width=700 border=1>";
echo "<tr><th>StudentID</th><th>StudentName</th><th>ProgrammeName</th></tr>";

while($row = mysql_fetch_array ($result))

{

echo "<tr><td>";
echo $row['StudentID'];
echo "</td><td>";
echo $row['StudentName'];
echo "</td><td>";
echo $row['ProgrammeName'];
echo "</td></tr>";

}

echo "</table>";
mysql_close($con);

?>

    how to create two dropdown menu linking each other?

    in Javascript Help

    Posted

    Hi guys, i have to create a drop-down menu for database data,  when i select one data  from that then next drop-down menu has to create automatically with related data of  the 1st selection.

    could anyone please give me some help to do this?

     

    NB: i have 3 tables student(sid, sname), course (cid, cname) , take (sid, cid) so if i select one of the cid then another drop down list  has to create automatically with sid who is taking the cid .

    cross tab query in mysql

    in MySQL Help

    Posted

    hi guys, i am  struggling with cross tab query in mysql. if you have any idea could you please help me? basically i have data into the table like...

     

    cid | Q# | marks

    c1 | 1| 50

    c1 | 2 | 50

    c1 | 3 | 50

     

    but i need to be the table like...

     

    cid | Q1 | Q2 | Q3

    c1 | 50 | 50 | 50

    inserting data from drop-down list & text field

    in PHP Coding Help

    Posted

    basically im displaying a column field as drop-down list.

    then ask enter No of question'

    (if i enter 2 then 2 text field will be created automatically)

    after i enter data into it.

    then all data have to be stored into another table.

    Eg:

    cid, q#,marks

    c1,1,50

    c1,2,50.

     

    here evrything has been displayed well but when i click submit, the data doesn't insert into table

     

     

    inserting data from drop-down list & text field

    in PHP Coding Help

    Posted

    Hi guys, basically here pull out the data from database then creating taxt field automatically and submit into anther table. everything works fine but data not inserting in to the table. could you guys check my code please?

     

     

     <?php

$con = mysql_connect("localhost","root","");

mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error());
?>
<?php
$result  = mysql_query("SELECT * FROM course") or trigger_error('MySQL error: ' . mysql_error());


echo '<select name ="cid[]">';

while($row = mysql_fetch_array($result))
{

    echo '<option value="' . $row['CourseID'] . '">'. $row['CourseID'] .'</option>';
}

echo '</select>';
//------------------
?>
   
<?php
if(!empty($_POST["submit"]))
{
$value = empty($_POST['question']) ? 0 : $_POST['question'];
?>
<form name="form1" method="post" action="result.php">
<?php


for($i=0;$i<$value;$i++)
{
  echo 'Question NO: <input type="text" name="qno[]" size="2" maxlength="2" class="style10">    
        Enter Marks: <input type="text" name="marks[]" size="3" maxlength="3" class="style10"><br>';
  }


?>
  <label> <br />
  <br />
  <input type="submit" name="Submit" value="Submit" class="style10">
  </label>
</form>
<?php
}
else{
?>

<form method="POST" action="#">



<label>
<span class="style10">Enter the Number of Question</span>
<input name="question" type="text" class="style10" size="2" maxlength="2"> 
</label>
<input name="submit" type="submit" class="style10" value="Submit">


</form>
<?php }?>

     

     

    result.php

    <?php


$con = mysql_connect("localhost","root","") or die('Could not connect: ' . mysql_error());
mysql_select_db("uni",$con) or die('Could not connect: ' . mysql_error());

foreach ($_POST['cid'] as $c) {$cid [] = $c;}
foreach($_POST['qno'] as $q){$qno[] = $q;}
foreach($_POST['marks'] as $m){$marks[] = $m;}
$ct = 0;
for($i=0;$i<count($qno);$i++)
{

$sql="INSERT INTO examquesion (CourseID,QuesionNo,MarksAllocated) VALUES('$cid[$i]','$qno[$i]','$marks[$i]')";
mysql_query($sql,$con) or die('Error: ' . mysql_error());
$ct++;
}
echo "$ct record(s) added";
mysql_close($con)
?>

     

    create text field dynamically!

    in PHP Coding Help

    Posted

    i did change as u said but still error.... im bit new & stupid with php things. :) could you check this for me pls?

    {
echo "Question NO:";
echo '<input type="text" name="qno[]">'."\t";

echo "Enter Marks:";
echo '<input type="text" name="marks[]">'."\t";
echo "<br>";
}

    $sql="INSERT INTO cell (QNO,MARKS)
VALUES
('$_POST['qno']','$_POST['marks']')";

     

    create text field dynamically!

    in PHP Coding Help

    Posted

    Hi guys i have to create text field & enter data and store in the data base. here im able to create text field but couldn't insert the data. so could anyone please check this code for me.

     

    <body>
<form method="POST" action="cell.php">
Enter the number of question
<input type="text" name="question">
<input type="submit">




  <?php
$value=$_POST['question'];

for($i=0;$i<$value;$i++)
{
echo "Question NO:";
echo '<input type="text" name="qno">'."\t";

echo "Enter Marks:";
echo '<input type="text" name="marks">'."\t";
echo "<br>";
}
?>
</form>
<form name="form1" method="post" action="cellresult.php">
  <label>
  <input type="submit" name="Submit" value="Submit">
  </label>
</form>

</body>

     

     

     

    cellresult.php

    <body>
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("test", $con);

$sql="INSERT INTO cell (QNO,MARKS)
VALUES
('$_POST[qno]','$_POST[marks]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

mysql_close($con)
?>
</body>

     

    check existing data insert into the table?

    in PHP Coding Help

    Posted

    Hi guys, im inserting data into the table using drop-down list & multi select list,well it works very well. but i need to make sure i should not insert same StudentID & CourseID twice. here my code for you could anyone tell me pls where should i write code to check existing data?

    <?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }


  mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error());
$result  = mysql_query("SELECT * FROM student") or trigger_error('MySQL error: ' . mysql_error());


echo '<select name="sid">';

while($row = mysql_fetch_array($result))
{

    echo '<option value="' . $row['StudentID'] . '">' . $row['StudentName'] . '</option>';
}

echo '</select>';

// ----------------

?>
</div>
<div class="style41" id="Layer7">

<?php
$result  = mysql_query("SELECT * FROM course") or trigger_error('MySQL error: ' . mysql_error());


echo '<select name ="cid[]" multiple="multiple" size="10">';

while($row = mysql_fetch_array($result))
{
    echo '<option value="' . $row['CourseID'] . '">' . $row['CourseName'] . '</option>';
}

echo '</select>';
mysql_close($con);
?>

     

     

    ------------------------------------

    <?php

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

  mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error());
   
if (!empty($_POST['sid']) && !empty($_POST['cid'])) 
{
    $ct = 0;
    $student = $_POST['sid'];
    foreach ($_POST['cid'] as $key => $course)
    {
     $sql = "INSERT INTO take (StudentID, CourseID) VALUES('".mysql_real_escape_string($student)."',
 '".mysql_real_escape_string($course)."')";
     $query = mysql_query($sql) or trigger_error('MySQL error: ' . mysql_error());
     if (mysql_affected_rows() > 0){$ct++;}
     }
    echo $ct . ' rows added.';
}  

mysql_close($con);

?>

     

     

    anyone check this code plss!

    in PHP Coding Help

    Posted

    still i get error msg. im bit new & stupid with php. so i just post all of my coding so could any1 check this for me plss

    <?php

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

  mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error());
   try
   {
$sql   = "SELECT StudentID,CourseID FROM take WHERE StudentID =" . $_POST['sid'] . " AND CourseID =" .$_POST['cid'] ;
    $query = mysql_query($sql) or trigger_error('MySQL error: ' . mysql_error());

    if (mysql_num_rows($query) > 0)
    {
        echo'StudentID already taken';
    }
}
if (!empty($_POST['sid']) && !empty($_POST['cid'])) //-------> error line
{
    $ct = 0;
    $student = $_POST['sid'];
    foreach ($_POST['cid'] as $key => $course)
    {
     $sql = "INSERT INTO take (StudentID, CourseID) VALUES('".mysql_real_escape_string($student)."','".mysql_real_escape_string($course)."')";
     $query = mysql_query($sql) or trigger_error('MySQL error: ' . mysql_error());
     if (mysql_affected_rows() > 0){$ct++;}
     }
    echo $ct . ' rows added.';
}  

mysql_close($con);

?>

     

    anyone check this code plss!

    in PHP Coding Help

    Posted 

    try
   {
$sql   = "SELECT StudentID,CourseID FROM student,course WHERE StudentID =" . $_POST['sid']  AND CourseID =".$_POST['cid'] ; //-----> error line

    $query = mysql_query($sql) or trigger_error('MySQL error: ' . mysql_error());

    if (mysql_num_rows($query) > 0)
    {
        throw new Exception('StudentID already taken');
    }
}

     

    NB: basicaly im inserting data into a table from another two table field.

    could anyone check this code

    in PHP Coding Help

    Posted 

    <body>

<form  method="post" action="new.php"  name="submitted"   /> 
  <label>TYPE: 
  <select name="field"> 
    <option value ="sid">StudentID</option> 
    <option value ="sname">StudentName</option> 
  </select> 
  </label> 
  <label>WORD: 
  <input type="text" name="searchword" /> 
  </label> 
  
  <input type="submit" name ="submitted" /> 
  
</form> 

<?php 
if (isset($_POST['submitted'])){ 
$searchword = htmlentities(addslashes($_POST['searchword']));

$con = mysql_connect("localhost","root",""); 

mysql_select_db("uni", $con) or trigger_error('MySQL error: ' . mysql_error());  

$field= $_POST['field'];
$searchword = $_POST['searchword'];


$result = mysql_query("SELECT* FROM student WHERE $field ='$searchword'") or trigger_error('MySQL error: ' . mysql_error());  

//$result = mysql_query($query); 
//$num_rows = mysql_num_rows($result); 

//echo"num_rows results found."; 
echo"<table>"; 
echo "<tr><th>StudentID</th><th>StudentName</th></tr>"; 
while($row = mysql_fetch_array($result)){ 

echo "<tr><td>"; 
echo "$row ['field']"; 
echo "</td><td>"; 
echo "$row ['searchword']"; 
echo "</td></tr>"; 
} 

echo"</table>"; 


} 

mysql_close($con); 

?>
</body>
</html>

     

    Hi guys, i jst made some correction but still some error on it. could you pls check this code for me.

     

    NB: my error msg.....

    Notice: MySQL error: Unknown column 'sid' in 'where clause' in C:\wamp\www\new.php on line 37

     

    Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\new.php on line 45

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