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How to get input field value using PHP
(7 answers)
Closed 6 years ago.
I have follow the tutorial of it where i want to update my database using two php files.
<?php
while($row = mysqli_fetch_array($records))
{
echo "<tr><form action =update.php method=post>";
echo "<td><input type=text name=Cname value='".$row['CustomerName']."'></td>";
echo "<td><input type=number name=size min=1 value='".$row['TableSize']."'></td>";
echo "<td><input type=date name=Adate value='".$row['DateA']."'></td>";
echo "<td><input type=time name=Atime value='".$row['TimeA']."'></td>";
echo "<td><input type=tel name=phonenumber value='".$row['PhoneNumber']."'></td>";
echo "<input type=hidden name=id value='".$row['TableID']."'>";
echo "<td><input type=submit>";
echo"</form></tr>";
}
?>
this is what i use for the first php file
as for the update.php:
<?php
$con = mysqli_connect('127.0.0.1','root','');
mysqli_select_db($con,'restaurant');
$sql = "UPDATE addtable SET CustomerName='$_POST[Cname]', TableSize='$_POST[size]', DateA='$_POST[Adate]',TimeA='$_POST[Atime]',PhoneNumber='$_POST[phonenumber]', WHERE TableID=$_POST[id]";
if(mysqli_query($con,$sql))
header("refresh:1; url=AssignBooking.php");
else
echo "Not Update";
?>
but the $sql line just doesn't work as it says that
Undefined index: Cname and other indexes too.
put quotes outside the post variable:
$sql = "UPDATE addtable SET CustomerName='".$_POST['Cname']."', TableSize='".$_POST['size']."', DateA='".$_POST['Adate']."',TimeA='".$_POST['Atime']."',PhoneNumber='".$_POST['phonenumber']."', WHERE TableID=".$_POST['id'];
According to your code put the name attributes value ' single quote.
<?php
while($row = mysqli_fetch_array($records))
{
echo "<tr><form action =update.php method=post>";
echo "<td><input type=text name='Cname' value='".$row['CustomerName']."'></td>";
echo "<td><input type=number name='size' min=1 value='".$row['TableSize']."'></td>";
echo "<td><input type=date name='Adate' value='".$row['DateA']."'></td>";
echo "<td><input type=time name='Atime' value='".$row['TimeA']."'></td>";
echo "<td><input type=tel name='phonenumber' value='".$row['PhoneNumber']."'></td>";
echo "<input type=hidden name="id" value='".$row['TableID']."'>";
echo "<td><input type=submit>";
echo"</form></tr>";
}
?>
Put quotes accordingly
UPDATE addtable SET CustomerName='".$_POST['Cname']."',TableSize='".$_POST['size']."', DateA='".$_POST['Adate']."',TimeA='".$_POST['Atime']."',PhoneNumber='".$_POST['phonenumber']."' WHERE TableID=$_POST['id'];
Related
I am trying to make a simple program where the user can multiply their inputs. But it always return a 0.
I already did:
$price = (int)$_POST['txtPrice'];
$quan = (int)$_POST['quantity'];
$ans = $quan * $price;
echo "$ans";
But returns 0 always.
This is the page where all the textboxes came from:
echo "<tr>";
echo "<td><input type='text' name='quantity' placeholder='How Many?'></td>";
echo "<td><input type='text' name='txtName' value='$name'></td>";
echo "<td><input type='text' name='txtPrice' value='$price'></td>";
Using your same code and just adding the form tag and a submit works for me:
<?php
if(!empty($_POST)) {
print_r($_POST);
$price = (int)$_POST['txtPrice'];
$quan = (int)$_POST['quantity'];
$ans = $quan * $price;
echo "$ans<br>";
}
echo "<tr>";
echo "<form action='index.php' method='post'>";
echo "<td><input type='text' name='quantity' placeholder='How Many?'> </td>";
echo "<td><input type='text' name='txtName' value='$name'></td>";
echo "<td><input type='text' name='txtPrice' value='$price'></td>";
echo "<input type='submit'>";
echo "</form>";
Execute it with php -S 127.0.0.1 index.php, $ans contains the value from the multiplication.
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 have an array of Textbox, by default they are all disabled... Some Textbox are empty some are not based on the return values of my query.
while ($imps_row1 = sqlsrv_fetch_array($stmt_line_util3,SQLSRV_FETCH_ASSOC)) {
echo "<tr>";
echo "<td><input type='text' disabled='disabled' class='txtDis' name='txt1[]' value='".$imps_row1['qualified_borrower']."'></td>";
echo "<td><input type='text' disabled='disabled' class='txtDis' name='txt2[]' value='".$imps_row1['allowed_borrower']."'></td>";
echo "</tr>";
}
Is there any way that the moment the page loads, all the textbox that are not empty will not be disabled?
Something this might work using just php:
while ($imps_row1 = sqlsrv_fetch_array($stmt_line_util3,SQLSRV_FETCH_ASSOC))
{
echo "<tr>";
echo "<td><input type='text' ".(empty($imps_row1['qualified_borrower'])?"disabled='disabled'":'')." class='txtDis' name='txt1[]' value='".$imps_row1['qualified_borrower']."'></td>";
echo "<td><input type='text' ".(empty($imps_row1['allowed_borrower'])?"disabled='disabled'":'')." class='txtDis' name='txt2[]' value='".$imps_row1['allowed_borrower']."'></td>";
echo "</tr>";
}
Just uses a ternary statement to check if the variable you are echo'ing to the value is empty and if so echo's "disabled...".
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='$result[\"quiz_ID\"]'>";
For some reason when I call a function to create a form, the drop down menus aren't sticky and the browser forces users to click into the first text field and tab through the rest. It won't let them mouse through the fields. This is only happening in FF, not IE or Chrome. The forms I'm including are just basic html and only php pages I include are doing this.
Here is one function:
function addNoteUI($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.