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june_c21

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  1. june_c21
    i got this error Parse error: syntax error, unexpected $end in /var/www/Web/RCA/rca.php on line 75 where it is a blank line . can someone tell me why and it is my code is wrong? i am new with php , please guide me. thanks in advance <style type="text/css"> <!-- .style2 { font-family: Times News Roman; font-size: 11px; } body { background-color: #CCFFFF; } .style6 { font-family: Arial; font-size: 14px; font-weight: bold; color: #FF0000; } .style5 { font-family: Arial; font-size: 12px; font-weight: bold; color: #000099; } .style10 {font-family: Times News Roman; font-size: 12px; } --> </style> <body> <form id="form1" name="form1" method="post" action="logout.php"> <label><div align="center"> <p class="style6">Root Cause Analysis </p> <p class="style2"><a href="logout.php">Logout</a></p> <p align="center" class="style10"><a href="home1.html">Home</a></p> </div> <div align="center"></div> </form> <?php $host = "localhost"; $user = "root"; $password = "admin"; $dbase = "rca"; $GF = $_POST ['GF']; $Year = $_POST ['Year']; if ($GF =='1' && $Year== '2006') { $dblink = mysql_connect($host,$user,$password); mysql_select_db($dbase,$dblink); echo "<table border=3 align=center width=100%>"; echo "<tr><td align=center><span class=style5>Reg ID.</td></span> <td align=center><span class=style5>Title</td></span> <td align=center><span class=style5>Incident Date</td></span> <td align=center><span class=style5>Fault Report ID</td></span> <td align=center><span class=style5>Report Received </td></span> <td align=center colspan=><span class=style5>Status </td></span> </tr>"; $query = "SELECT reg_no,title1,incident_date,fault_rep_id,secr_recd,status from report where unit='1' || unit='2' "; $result = mysql_query($query, $dblink); while($myrow = mysql_fetch_row($result)) { echo "<tr><td align=center><span class=style2>" . $myrow[0] . "</span</td><td align=center><span class=style2>" . $myrow[1] . "</span</td><td align=center><span class=style2>" . $myrow[2] . "</td></span><td><span class=style2>" . $myrow[3] . "</span></td><td align=center><span class=style2>" . $myrow[4] . "</span></td><td align=center><span class=style2>" . $myrow[5] . "</span></td></tr>"; } } else { echo "HELLO"; } echo "</table>"; ?>
  2. june_c21
    thanks barand. topic solved!!!!!
  3. june_c21
    now i redo the html file again <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> </head> <body> <form id="form1" name="form1" method="post" action="checking2.php"> <label>Name <input name="staff_no" type="text" id="staff_no" /> </label> <p> <label> <input type="submit" name="Submit" value="Submit" /> </label> </p> </form> </body> </html> php file <?php // echo " New Entry <a href=add.html>click here</a><br></br>"; $host = "localhost"; $user = "root"; $password = ""; $dbase = "claim"; $dblink = mysql_connect($host,$user,$password); mysql_select_db($dbase,$dblink); $staff_no = $_POST['staff_no']; echo "<table border=3 align=center width=90%>"; echo "<tr><td width=12% align=center><span class=style5>ID</td></span>"; echo "<td width=12% align=center><span class=style5>Month</td></span>"; echo "<td width=12% align=center><span class=style5>Start Date </td></span>"; echo "<td width=12% align=center><span class=style5>End Date</td></span>"; echo "<td align=center><span class=style5>Name</td></span>"; echo "<td align=center><span class=style5>Staff No</td></span>"; echo "<td align=center><span class=style5>Details</td></span>"; echo "<td align=center><span class=style5>Amount</td></span>"; echo "<td align=center><span class=style5>GL code</td></span>"; echo "<td width=12% align=center><span class=style5>Bank </td></span>"; echo "<td align=center ><span class=style5>Account No. </td></span>"; echo "<td align=center ><span class=style5>Total </td></span>"; echo "<td align=center colspan=3><span class=style5>Action </td></span></tr>"; $query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no,details,amount,gl_code,user.bank,user.acc_no FROM user,report1 WHERE report1.staff_no = '$staff_no'"; $result = mysql_query($query, $dblink); while($myrow = mysql_fetch_row($result)) { echo "<tr><td><span class=style2>" . $myrow[0] . "</span</td>"; echo "<td><span class=style2>" . $myrow[1] . "</span</td>"; echo "<td><span class=style2>" . $myrow[2] . "</span</td>"; echo "<td><span class=style2>" . $myrow[3] . "</td></span>"; echo "<td><span class=style2>" . $myrow[4] . "</span></td>"; echo "<td><span class=style2>" . $myrow[5] . "</span></td>"; echo "<td><span class=style2>" . $myrow[6] . "</span></td>"; echo "<td><span class=style2>" . $myrow[7] . "</span></td>"; echo "<td><span class=style2>" . $myrow[8] . "</span></td>"; echo "<td><span class=style2>" . $myrow[9] . "</span></td>"; echo "<td><span class=style2>" . $myrow[10] . "</span></td>"; $query1 = "SELECT SUM(amount) FROM report1 where staff_no = '$myrow[5]'"; $result1 = mysql_query($query1, $dblink); while ($myrow1 = mysql_fetch_row($result1)) { echo "<td><span class=style2>" .$myrow1[0]. "</span></td>"; } } echo "</table>"; ?> why when i input one particular staff_no, it display blank when i click on submit button
  4. june_c21
    try already. no error
  5. june_c21
    <?php //$id=$_GET["id"]; $staff_no = $_POST['staff_no']; $host = 'localhost'; $user = 'root'; $password = ''; $dbase = 'claim'; $dblink = mysql_connect($host,$user,$password); mysql_select_db($dbase,$dblink); $sql="SELECT report1.staff_no, user.name FROM report1 JOIN user ON report1.staff_no = user.staff_no WHERE report1.staff_no = '.$staff_no.'"; $result = mysql_query($sql); echo "<table border='1'> <tr> <th>Staff No</th> <th>Name</th> </tr>"; while($row = mysql_fetch_array($result)) { $report1_staffno = $row['report1.staff_no']; $username = $row['user.name']; echo "<tr>"; echo "<td>$report1_staffno</td>"; echo "<td>$username</td>"; echo "</tr>"; } echo "</table>"; ?> this is the user input code <html> <head> <script src="selectuser.js"></script> </head> <body><form id="form1" name="form1" method="post" action="checking.php"> <p><strong>KEV Tool Store - Loan Transaction Page</strong></p> <table width="497" border="0"> <tr> <td width="284">Choose</td> <td width="203"><select name="staff_no" onChange="showUser(this.value)"> <?php $host = 'localhost'; $user = 'root'; $password = ''; $dbase = 'claim'; $dblink = mysql_connect($host, $user, $password); mysql_select_db($dbase, $dblink); $query = "SELECT staff_no FROM user ORDER BY staff_no "; $result = mysql_query($query); while($row = mysql_fetch_array($result)) { $staffno = $row['staff_no']; echo "<option value=\"$staffno\">$staffno</option>"; } ?> </select></td> </tr> <tr> <td colspan="2"><p> <div id="txtHint"><b>info will be listed here.</b></div> </p> </td> </tr> </table> <p> </p> <p> <input type="submit" name="Submit" value="Submit"> </p> </form></body> </html> this whole code just can't get me the name and staff no when user input their staff_no . why? i been trying for 3 days....
  6. june_c21
    barand, why it came out blank instead of showing the staff_no and name? anything i miss out?
  7. june_c21
    report1 is a table name user is another table name i need to match report1.staff_no (databse) with user input staff_no and then based on both table got staff_no, i will take the name from user table that match with the user input staff_no
  8. june_c21
    ok, i test and this code got error SELECT report1.staff_no, user.name FROM report1,user WHERE report1.staff_no = '$staff_no' when i replace sk200023 into $staff_no all the name in the database will came out and they auto change their staff_no into sk200023. why?
  9. june_c21
    sorry .. is still doesn't work. it came out blank only
  10. june_c21
    why my code can't execute? when i execute, it output all the staff_no, name instead of specific staff_no only <?php //$id=$_GET["id"]; $staff_no = $_POST['staff_no']; $host = 'localhost'; $user = 'root'; $password = ''; $dbase = 'claim'; $dblink = mysql_connect($host,$user,$password); mysql_select_db($dbase,$dblink); $sql="SELECT report1.staff_no, user.name FROM report1,user WHERE report1.staff_no = '".$staff_no."'"; $result = mysql_query($sql); echo "<table border='1'> <tr> <th>Staff No</th> <th>Name</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row[0] . "</td>"; echo "<td>" . $row[1] . "</td>"; echo "</tr>"; } echo "</table>"; ?>
  11. june_c21
    what do you mean? sorry... i don't understand.
  12. june_c21
    hi, can someone tell me what is the error of this code. i try it many times but in the drop down menu there is no data that i select from the database . Select a User: <select name="staff no " onChange="showUser(this.value)"> <?php $host = 'localhost'; $user = 'root'; $password = ''; $dbase = 'claim'; $dblink = mysql_connect($host, $user, $password); mysql_select_db($dbase, $dblink); $query = "SELECT staff_no FROM user ORDER BY staff_no "; $result = mysql_query($query); while($myrow = mysql_fetch_row($result)) { echo "<option value=\"" . $myrow[0] . "\">" ; echo "</option>";}
  13. june_c21
    BUMP
  14. june_c21
    no .. it still doesnt work. please help and guide me more. i am new. thanks in advances
  15. june_c21
    but that haven't solve my problem. i need to output the user that have the same staff no not all the user. i wrote SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no, details, amount, gl_code, user.bank, user.acc_no FROM user, report1 WHERE report1.staff_no = '$name' but it output all the staff.
  16. june_c21
    how to do so?
  17. june_c21
    it came out all the username from database and changed their stuff_no into same staff_no. (250 records)
  18. june_c21
    the result still the same.... can someone please help me..... brian, thanks for your code... but it doesn't work..
  19. june_c21
    brian, did u changed anything in the code?
  20. june_c21
    ok, first this is my html coding where user can chose the staff_no and click on submit. once submit, it will find in the database the same staff_no that match the one user select. if it is yes, it will display the records base on the staff_no . <html> <head> <script src="selectuser.js"></script> </head> <body><form id="form1" name="form1" method="post" action="checking.php"> <p><strong>KEV Claim</strong></p> <table width="497" border="0"> <tr> <td width="284">Claim</td> <td width="203"><select name="name" id="name" onChange="showUser(this.value)"> <?php $host = 'localhost'; $user = 'root'; $password = 'admin'; $dbase = 'claim'; $dblink = mysql_connect($host,$user,$password); mysql_select_db($dbase,$dblink); $query = "SELECT staff_no, name from user order by name "; $result = mysql_query($query); while($myrow = mysql_fetch_row($result)) { echo "<OPTION VALUE=\"" . $myrow[0] . "\">" . $myrow[1]."</OPTION>"; } ?> </select></td> </tr> </table> <p> </p> <p> <input type="submit" name="Submit" value="Submit"> </p> </form></body> </html> <?php $host = "localhost"; $user = "root"; $password = "admin"; $dbase = "claim"; $dblink = mysql_connect($host,$user,$password); mysql_select_db($dbase,$dblink); echo "<table border=3 align=center width=90%>"; echo "<tr><td>ID</td>"; echo "<td>Month</td>"; echo "<td>Start Date </td>"; echo "<td>End Date</td>"; echo "<td>Name</td>"; echo "<td>Staff No</td>"; echo "<td>Details</td>"; echo "<td>Amount</td>"; echo "<td>GL code</td>"; echo "<td>Bank </td>"; echo "<td>Account No. </td><tr>"; $name = $_POST['name']; $query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no, details, amount, gl_code, user.bank, user.acc_no FROM user, report1 WHERE report1.staff_no = '$name'"; $result = mysql_query($query, $dblink); while($myrow = mysql_fetch_array($result)) { echo "<tr><td>" . $myrow[0] . "</td>"; echo "<td>" . $myrow[1] . "</td>"; echo "<td>" . $myrow[2] . "</td>"; echo "<td>" . $myrow[3] . "</td>"; echo "<td>" . $myrow[4] . "</td>"; echo "<td>" . $myrow[5] . "</td>"; echo "<td>" . $myrow[6] . "</td>"; echo "<td>" . $myrow[7] . "</td>"; echo "<td>" . $myrow[8] . "</td>"; echo "<td>" . $myrow[9] . "</td>"; echo "<td>" . $myrow[10] . "</td></tr>"; } echo "</table>"; ?>
  21. june_c21
    Brian, the result still the same. all data came out when i click on submit button.
  22. june_c21
    it still the same. all the data will come out instead of match report1.staff_no = '$name' only. why?
  23. june_c21
    now there is no error. thanks for all the help . but why when i wrote $name = $_POST['name']; $query = "SELECT report1.id,month,start_date,end_date,user.name,report1.staff_no,details,amount,gl_code,user.bank,user.acc_no FROM user,report1 WHERE report1.staff_no = '$name' "; $result = mysql_query($query, $dblink); while($myrow = mysql_fetch_row($result)) { echo "<tr><td>" . $myrow[0] . "</td>"; echo "<td>" . $myrow[1] . "</td>"; echo "<td>" . $myrow[2] . "</td>"; echo "<td>" . $myrow[3] . "</td>"; echo "<td>" . $myrow[4] . "</td>"; echo "<td>" . $myrow[5] . "</td>"; echo "<td>" . $myrow[6] . "</td>"; echo "<td>" . $myrow[7] . "</td>"; echo "<td>" . $myrow[8] . "</td>"; echo "<td>" . $myrow[9] . "</td>"; echo "<td>" . $myrow[10] . "</td></tr>"; } it came out all the data instead of where the staff_no = $name only?
  24. june_c21
    thanks rohan but now the parse error: syntax error, unexpected T_VARIABLE is at line 26 which is this line $query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no,details,amount,gl_code,user.bank,user.acc_no FROM user,report1 WHERE user.name = "$name"";
  25. june_c21
    can someone tell me what wrong with my wrong code? i get this error "Parse error: syntax error, unexpected T_VARIABLE line 2" <? php $host = 'localhost'; $user = 'root'; $password = 'admin'; $dbase = 'claim'; $dblink = mysql_connect($host,$user,$password); mysql_select_db($dbase,$dblink); echo "<table border=3 align=center width=90%>"; echo "<tr><td>ID</td>"; echo "<td>Month</td>"; echo "<td>Start Date </td>"; echo "<td>End Date</td>"; echo "<td>Name</td>"; echo "<td>Staff No</td>"; echo "<td>Details</td>"; echo "<td>Amount</td>"; echo "<td>GL code</td>"; echo "<td>Bank </td>"; echo "<td>Account No. </td><tr>"; $name = $_GET['name']; $query = "SELECT report1.id, month, start_date, end_date, user.name, report1.staff_no,details,amount,gl_code,user.bank,user.acc_no FROM user,report1 WHERE user.name = "$name"; $result = mysql_query($query, $dblink); while($myrow = mysql_fetch_row($result)) { echo "<tr><td>" . $myrow[0] . "</td>"; echo "<td>" . $myrow[1] . "</td>"; echo "<td>" . $myrow[2] . "</td>"; echo "<td>" . $myrow[3] . "</td>"; echo "<td>" . $myrow[4] . "</td>"; echo "<td>" . $myrow[5] . "</td>"; echo "<td>" . $myrow[6] . "</td>"; echo "<td>" . $myrow[7] . "</td>"; echo "<td>" . $myrow[8] . "</td>"; echo "<td>" . $myrow[9] . "</td>"; echo "<td>" . $myrow[10] . "</td></tr>"; } echo "</table>"; ?>
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