Please help, each row in the table has save button for editing/adding content. I have 3 textboxes in each row,PersoninCharge,PIC_Comments and Status. The user can add/edit these textboxes whenever they click the save button on that particular row. The problem is, whenever I add/edit data in one row, the save button can't read what particular row I edited.
The save button is perfectly running if I searched the invoice number first, but what I want is to directly edit/add the data in each row without searching it first.
Here's the code:
For the table:
<?php
if (mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_array($result))
{
echo"<tr class=output2>";
echo "<td>$row[1]</td>";
echo "<td>$row[2]</td>";
echo "<td>$row[3]</td>";
echo "<td>$row[4]</td>";
echo "<td>$row[5]</td>";
echo "<td>$row[6]</td>";
echo "<td>$row[7]</td>";
echo "<td>$row[8]</td>";
echo "<td>$row[9]</td>";
echo "<td>$row[10]</td>";
echo "<td>$row[11]</td>";
echo "<td>$row[12]</td>";
echo "<td><input type='text' name='pic' value='$row[17]'></td>";
echo "<td><input type='text' name='comt' value='$row[18]'></td>";
echo "<td><input type='text' name='stat' value='$row[19]'></td>";
echo "<td><form name='update' method='POST'><input type='submit' name='save_btn' value='SAVE' style='font-size:1em;'/></form></td>";
echo "<td><input type='hidden' name='idtxt' value='$row[0]'/></td>";
echo "</tr>";
}
}
else
{
echo '<h3>No result found! </h3><br>';
}
$con->close();
For Save button:
if(isset($_POST['save_btn']))
{
$query2="UPDATE invalid_invoice SET UpdateBy='".$_SESSION['login_user']."', UpdateDateTime=NOW(), PersoninCharge='".$_POST['pic']."', PIC_Comments='".$_POST['comt']."', Status='".$_POST['stat']."' WHERE ID='".$_POST['idtxt']."'";
$con->query($query2);
$con->close();
echo '<h3 class="datasuccess">Data successfully added!</h3>';
}
I have edited your form. please have a look
<?php
if (mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_array($result))
{
echo"<form name='update' method='POST'><tr class=output2>";
echo "<td>$row[1]</td>";
echo "<td>$row[2]</td>";
echo "<td>$row[3]</td>";
echo "<td>$row[4]</td>";
echo "<td>$row[5]</td>";
echo "<td>$row[6]</td>";
echo "<td>$row[7]</td>";
echo "<td>$row[8]</td>";
echo "<td>$row[9]</td>";
echo "<td>$row[10]</td>";
echo "<td>$row[11]</td>";
echo "<td>$row[12]</td>";
echo "<td><input type='text' name='pic' value='$row[17]'></td>";
echo "<td><input type='text' name='comt' value='$row[18]'></td>";
echo "<td><input type='text' name='stat' value='$row[19]'></td>";
echo "<td><input type='submit' name='save_btn' value='SAVE' style='font-size:1em;'/></td>";
echo "<td><input type='hidden' name='idtxt' value='$row[0]'/></td>";
echo "</tr></form>";
}
}
else
{
echo '<h3>No result found! </h3><br>';
}
$con->close();
Related
I am trying to create a php page where the materials from database are populated. Users should be able to enter the quantity next to the item they wish to order and I have created a qty text field for this
<?php
session_start();
include("db.php");
$pagename="Order Material";
echo "<html>";
echo "<title>".$pagename."</title>";
echo "<h2>".$pagename."</h2>";
include ("detectlogin.php");
echo "<link rel=stylesheet type=text/css href=mystylesheet.css>";
$sql="select * from material";
$result=mysqli_query($con, $sql) or die(mysqli_error($con));
echo "<table border=1>";
echo "<tr>";
echo "<th>Material Name</th>";
echo "<th>Material Description</th>";
echo "<th>Toxicity Level</th>";
echo "</tr>";
while ($arraymaterials=mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>".$arraymaterials['materialName']."</td>";
echo "<td>".$arraymaterials['materialDescrip']."</td>";
echo "<td>".$arraymaterials['materialToxicity']."</td>";
echo "<td>Enter Quantity</td>";
echo "<td><input type=text name=qty value=qty size=5></td>";
echo "<form action=request_material.php method=post>";
echo "<input type=hidden name=materialcode value=".$arraymaterials['materialCode'].">";
echo "<td><input type=submit value='Request'></td>";
echo "</form>";
echo "</tr>";
}
echo "</table>";
?>
However, I cannot successfully post the value of qty on to the next page even though I have $qty=$_POST['qty']; on my request_material.php. Do you know why this value entered in the qty field cannot be posted onto the request_material.php page? Do I need a session variable?
thanks
Because input tag name="qty" is outside the form tag
echo "<td><input type=text name=qty value=qty size=5></td>";// outside form tag
echo "<form action=request_material.php method=post>";
echo "<input type=hidden name=materialcode value=" . $arraymaterials['materialCode'] . ">";
echo "<td><input type=submit value='Request'></td>";
echo "</form>";
You need to add it inside your form tag as
echo "<form action=request_material.php method=post>";
echo "<td><input type=text name=qty value=qty size=5></td>";// add inside it
echo "<input type=hidden name=materialcode value=".$arraymaterials['materialCode'].">";
echo "<td><input type=submit value='Request'></td>";
echo "</form>";
Please try this: I have updated the code:
echo "<table border=1>";
echo "<tr>";
echo "<th>Material Name</th>";
echo "<th>Material Description</th>";
echo "<th>Toxicity Level</th>";
echo "</tr>";
if(mysqli_num_rows($result)>0)
{
echo "<form action=request_material.php method=post>";
while ($arraymaterials=mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>".$arraymaterials['materialName']."</td>";
echo "<td>".$arraymaterials['materialDescrip']."</td>";
echo "<td>".$arraymaterials['materialToxicity']."</td>";
echo "<td>Enter Quantity</td>";
echo "<td><input type='text' name='qty[]' value='qty' size=5></td>";
echo "<input type=hidden name='materialcode[]' value=".$arraymaterials['materialCode'].">";
echo "<td><input type=submit value='Request'></td>";
echo "</tr>";
}
echo "</form>";
}
echo "</table>";
In request_material.php check value of $qty.It will be an array.for more details print_r($_POST) in request_material.php
I'm trying to design a control page for my website that allows admins to delete requests to use a 3D printer at my school. The page displays a query of the schedule from the MySQL database and puts a delete button next to every entry. However, I would like to link the information in each entry to its respective delete button and have no idea how to do that. You can see the page itself here: http://kepler.covenant.edu/~techclub/PHP/manage%20printer%20schedule.php and this is my code for displaying the information
if ($result = $mysqli->query($query)) {
// see if any rows were returned
if ($result->num_rows > 0) {
// yes
// print them one after another
echo "<table cellpadding=10 border=1>";
echo "<tr>";
echo "<th>Control</th>";
echo "<th>Student ID</th>";
echo "<th>Last Name</th>";
echo "<th>First Name</th>";
echo "<th>Date and Time</th>";
echo "<th>Description</th>";
echo "</tr>";
$key = 1;
while($row = $result->fetch_array()) {
echo '<form method="POST" name="deleterequest" action = "delete request.php">';
echo "<tr>";
echo "<td><input name='".$row[5]."'type='submit' value='Delete' ></td>";
echo "<td>".$row[0]."</td>";
echo "<td>".$row[1]."</td>";
echo "<td>".$row[2]."</td>";
echo "<td>".$row[3]."</td>";
echo "<td>".$row[4]."</td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
}
else {
// no
// print status message
echo "No upcoming prints found!";
}
// free result set memory
$result->close();
}
I can't figure out how to link the information in a row with the delete button that I put in the row to send to the delete request.php program (which I have no code for yet)
you should put hidden field that hold each record id.
As below
while($row = $result->fetch_array()) {
echo "<tr>";
echo "<td>";
echo '<form method="POST" name="deleterequest" action = "delete request.php">';
echo "<input name='recored_id' type='hidden' value='".$row['id']."' >";
echo "<input name='delete'type='submit' value='Delete' >";
echo "</form>";
echo "</td>";
echo "<td>".$row[0]."</td>";
echo "<td>".$row[1]."</td>";
echo "<td>".$row[2]."</td>";
echo "<td>".$row[3]."</td>";
echo "<td>".$row[4]."</td>";
echo "</tr>";
echo "</form>";
}
Now from your request.php page you should do something like this
$recored_id = (int) $_POST['recored_id'];
$sql = "DELETE FROM table_name WHERE recored_id='$recored_id'";
Don't forget to give this code some security validation
all the best,
You're going to need some hidden inputs, like so:
while($row = $result->fetch_array()) {
echo "<tr>";
echo '<td><form method="POST" name="deleterequest" action = "deleterequest.php">';
echo "<input type='hidden' name='val0' value='" . $row[0] . "'>";
echo "<input type='hidden' name='val1' value='" . $row[1] . "'>";
echo "<input type='hidden' name='val2' value='" . $row[2] . "'>";
echo "<input type='hidden' name='val3' value='" . $row[3] . "'>";
echo "<input type='hidden' name='val4' value='" . $row[4] . "'>";
echo "<input name='".$row[5]."'type='submit' value='Delete' >";
echo "</form></td>";
echo "<td>".$row[0]."</td>";
echo "<td>".$row[1]."</td>";
echo "<td>".$row[2]."</td>";
echo "<td>".$row[3]."</td>";
echo "<td>".$row[4]."</td>";
echo "</tr>";
}
You should change the name of these inputs though
Please refer to the image below:
http://i.stack.imgur.com/6hBPC.png
For instance, if a user clicks the button on the row which says "You have a quiz for math", the "Quiz ID" value of THAT row would then be passed to another PHP file.
Here's my current code:
<?php
$con=mysqli_connect("127.0.0.1", "root", "", "quizmaker");
if (mysqli_connect_errno($con))
{
echo "MySqli Error: " . mysqli_connect_error();
}
$now=date("m/d/Y");
$sql=mysqli_query($con,"SELECT * FROM quiz_query WHERE quiz_date='$now'");
$count=mysqli_num_rows($sql);
if($count>=1)
{
echo "<table border='1' width='50%'>";
echo "<form action='answer_quiz.php' method='post'>";
echo "<tr>
<td>You have a pending quiz!</td><td> </td><td> </td>
</tr>";
$number=1;
while($result=mysqli_fetch_array($sql))
{
echo "<tr>";
echo "<td>You have a quiz for " . $result['subject'] . "</td>";
echo "<td>Quiz ID: " .$result['quiz_ID']. "</td>";
echo "<td><input type='submit' name='button' id='button' value='Take Quiz'>";
echo "<input type='hidden' name='quiz[$number]' value='$result[quiz_ID]'>";
echo "</td>";
echo "</tr>";
$number++;
}
echo "</form>";
echo "</table>";
}
else
{
"You have no quiz! :D";
}
mysqli_close($con);
?>
Move this line:
echo "<form action='answer_quiz.php' method='post'>";
Inside of the while loop.
Also, change
echo "<input type='hidden' name='quiz[$number]' value='$result[quiz_ID]'>"
with
echo "<input type='hidden' name='quizId' value='$result[quiz_ID]'>"
Now, in answer_quiz.php you'll receive $_POST['quizId'] with the value you need.
Change your while to :
while( $row = $result->fetch_array(MYSQLI_ASSOC)){
echo $row['subject'];
}
You are forgetting quotes around your variable:
Instead of
echo "<input type='hidden' name='quiz[$number]' value='$result[quiz_ID]'>";
It should be
echo "<input type='hidden' name='quiz[$number]' value='$result[\"quiz_ID\"]'>";
I am just trying to get the values from a table, but for some reason GET isn't working for me, or I am doing something wrong. Here is how I create my table in one php file:
<?php
.
.
.
.
echo "<tr>";
echo "<td>Login ID</td>";
$j=1;
echo "<td><input name=$j type='text' value='$line[$j]' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Name</td>";
$j=2;
echo "<td><input name=$j type='text' value='$line[$j]' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Password</td>";
$j=3;
echo "<td><input name=$j type='text' value='$line[$j]' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Birthday</td>";
$j=4;
echo "<td><input name=$j type='text' value='$line[$j]' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Address</td>";
$j=5;
echo "<td><input name=$j type='text' value='$line[$j]' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Email</td>";
$j=6;
echo "<td><input name=$j type='text' value='$line[$j]' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Phone Number</td>";
$j=7;
echo "<td><input name=$j type='text' value='$line[$j]' size=20/></td>";
echo "</tr>";
?>
The names should be the number 1-7 right?
In another php file I attempt to access the values in those fields with he following code:
<?php
.
.
.
.
$login_id = $_GET['1'];
$name = $_GET['2'];
$pw = $_GET['3'];
$bday = $_GET['4'];
$address = $_GET['5'];
$email = $_GET['6'];
$phno = $_GET['7'];
echo "new: $login_id, $name, $pw, $bday, $address, $email, $phno";
?>
Here is what I end up getting back: new: , , , , , ,
So what is it that I am doing wrong? I can't seem to find anything wrong with my code. I know I should probably be using $_POST for the password.
<?php
echo "<form action='otherfile.php' method='get'><tr>";
echo "<td>Login ID</td>";
$j=1;
echo "<td><input name=$j type='text' value='$j' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Name</td>";
$j=2;
echo "<td><input name=$j type='text' value='$j' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Password</td>";
$j=3;
echo "<td><input name=$j type='text' value='$j' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Birthday</td>";
$j=4;
echo "<td><input name=$j type='text' value='$j' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Address</td>";
$j=5;
echo "<td><input name=$j type='text' value='$j' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Email</td>";
$j=6;
echo "<td><input name=$j type='text' value='$j' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Phone Number</td>";
$j=7;
echo "<td><input name=$j type='text' value='$j' size=20/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>Submit Data</td>";
echo "<td><input name='submitdata' type='submit' value='submit' size=20/></td>";
echo "</tr><form>";
?>
if you want submit then get the all value in otherfile.php file
try its working...
You have not put the <form> tag. Add tag in begining
<form action='another_page.php' method='get'>
and in the end write
<input type='submit' name='submit>
</form>
What it will do on submit it will forward the values to the another page.
<form name="login" method="get" action="youraction.php">
echo "<td><input name=$j type='text' value=$j /></td>";
</form>
youraction.php
---------------
<?php
if($_GET)
{
print_r($_GET);
}
?>
For some reason when I call a function to create a form, the drop down menus aren't sticky and the browser forces users to click into the first text field and tab through the rest. It won't let them mouse through the fields. This is only happening in FF, not IE or Chrome. The forms I'm including are just basic html and only php pages I include are doing this.
Here is one function:
function addNoteUI($keyword) {
echo "<div id='search_result_right'>";
echo "<center><div id='enter_note_header'>Assign a Salesperson</div></center><p>";
echo "<form id='response' action='notes_add.php' method='post'>";
echo "<label for='mod_num'>MOD Initials: <label>";
echo "<input type='text' name='mod_num' size='2' maxlength='4'><p>";
echo "<label for='sales_num'>Assigned to Sales Person: <label>";
echo "<input type='text' name='sales_num' size='2' maxlength='4'><p>";
echo "<input type='hidden' name='question_num' value='$keyword'>";
echo "<label for='response'>Note</label><br>";
echo "<textarea name='response' cols='30' rows='7 maxlength='510'></textarea><p>";
echo "<input type='submit' value='Assign'>";
echo "</form>";
echo "</div>";
Here is the other:
function changeDept() {
include 'ask_search.php';
echo "<div id='search_result'>";
echo "<form action='change_dept.php' method='post'>";
echo "<label for='current_num'>Enter the Question Number to be Changed: <label>";
echo "<input type='text' name='current_num' size='4'><p>";
echo "<label for='store'>Select New Store/Department: <label>";
echo "<select name='store'>";
echo "<option>Please Select</option>";
echo "<option value='Albany'>Sales (Albany Store)</option>";
echo "<option value='Saratoga'>Sales (Saratoga Store)</option>";
echo "<option value='Web Sales'>Sales (TaftFurniture.com)</option>";
echo "<option value='Financing'>Financing</option>";
echo "<option value='Customer Service'>Customer Service</option>";
echo "<option value='Delivery'>Delivery</option>";
echo "<option value='HR'>Human Resources</option>";
echo "<option value='Web Contact'>Website Comment</option>";
echo "<input type='submit' value='Change' id='dropdown'>";
echo "</select></form></div>";
}
Thanks in advance.
Your labels are not closing properly:
echo "<label for='mod_num'>MOD Initials: <label>";
Should be:
echo "<label for='mod_num'>MOD Initials: </label>";
Also, in the second example, you have an input inside the select. The input must be outside:
echo "<option value='Web Contact'>Website Comment</option>";
echo "<input type='submit' value='Change' id='dropdown'>";
echo "</select></form></div>";
Should be:
echo "<option value='Web Contact'>Website Comment</option>";
echo "</select>";
echo "<input type='submit' value='Change' id='dropdown'></form></div>";
And another one, you're not closing your P tags:
echo "<input type='text' name='mod_num' size='2' maxlength='4'><p>";
Should be:
echo "<p><input type='text' name='mod_num' size='2' maxlength='4'></p>";
Try to be more careful with your tags. Some browsers are more forgiving about malformed HTML, but others are not.