mtvaran

yes now it works. i had to take echo mysql_error(); off. previously i put it for error finding. thanks guys for your kind help.

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); ?>

i need to use the StudentID from take table

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

whats wrong with query and how should i change it? could you please tell me?

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); ?>

could anyone provide code for me how to search data using column name. example: if i have 2 columns called NAME,ID then i select NAME or ID and enter the keyword to search. would greatly appreciate your help!

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 .

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

no dear, still error. im bit new with php so the problem is i couldn't find error data easily.

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

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) ?>

sorry i do not get the array things. i know im bit stupid. so could anyone pls provide the code for rest of it?

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']')";

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>

you mean like this.... this also give me error..... $sql = "SELECT * FROM take WHERE StudentID =" . $_POST['sid'] . "AND CourseID =" . $_POST['cid']; if not could you please correct it for me. thanks

Can i write like this....? this gives me an error msg.... like Parse error: parse error in C:\wamp\www\finalresult.php on line 26 can anyone modify this for me? $sql = "SELECT StudentID FROM take WHERE StudentID =" . $_POST['sid'] "AND CourseID =" . $_POST['cid'];

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); ?>

this is the error msg.... Parse error: parse error, expecting `T_CATCH' in

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); ?>

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.

still error mate...... Notice: MySQL error: Unknown column 'sid' in 'where clause' in C:\wamp\www\new.php on line 39 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\new.php on line 47

<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

no dear, still some error. i dnt knw y could you please help me to sort it out? NB: as you know the thing is i have table (two field StudentID & StudentName) in my data base i need to search by keywords and display the data on the web page

no data displayed. only shows "error data" as i mentioned below... $result = mysqli_query($con,$query) or die ('error data'); //-----------> error line