I am trying to display the name retrieved from Database in html file . Referred couple of blogs, still I am not able to display the value
<?php
$result=mysql_query("SELECT * from `Data` where Emp_Id = $EmpID")
$info = mysql_fetch_array( $result );
$Emp_ID=$info['Emp_ID'];
<html><head>
<body>
<input name="emp_number" value=<?php echo $Emp_ID?>/>
<input name='emp_name' value="<?php echo (isset($Emp_ID) ? htmlspecialchars($Emp_ID) : ''); ?>" />
<input name='emp_name'<?php echo (isset($Emp_Name)) ? ('value = "'.$Emp_Name.'"') : "value = \"\""; ?>/>
</body></head></html>
?>
Tried three ways, but none of them worked.
You should use a template system.
Beside of that try this:
<?php
$result=mysql_query("SELECT * from `Data` where Emp_Id = $EmpID")
$info = mysql_fetch_array( $result );
$Emp_ID=$info['Emp_ID'];
?>
<html>
<body>
<input name="emp_number" value=<?php echo $Emp_ID; ?>/>
<input name='emp_name' value="<?php echo (isset($Emp_ID) ? htmlspecialchars($Emp_ID) : ''); ?>" />
<input name='emp_name'<?php echo (isset($Emp_ID)) ? ('value = "'.$Emp_ID.'"') : "value = \"\""; ?>/>
</body>
</html>
This is the third time this week I've seen a similar question to this.
Please search SO and look for related questions
$EmpID is never being set
Your code is vulnerable to SQL injection and your site is likely to soon get hacked
Your query has a discrepancy between Emp_ID and your returned result Emp_Id
You've got two input fields with the same name, so only one will be returned on Post/Get
$Emp_Name is never set anywhere (guessing it should come form info?)
Also on the third input, you should have a space before you echo your "value=" field
Looks like the Emp_ID in line three should be Emp_Id or the Emp_Id in line one should be Emp_ID.
Related
I am trying to increment project number based on the last entry. The the primary key PROJECTNOID auto-increments but is not the same format as the project number (Ex: PROJECTNOID = 1 and Project Number = 19000). I don't want this to be a dropdown box even though some of my code shows the opposite.
<?php
connect = mysqli_connect("**", "**", "**", "**");
$query4 = "SELECT PROJECTNOID, ProjectNumber FROM tblProjects ORDER BY
PROJECTNOID";
$result4 = mysqli_query($connect,$query4);
$options4 = "";
while($row4 = mysqli_fetch_row($result4);){
$options4 = $options4."<input value=$row4[0]$row4[1]</input>";
}
?>
Here is the html textbox:
<label for="txtfield">Project Number</label>
<!--<input type="text" id="reqtxtfield" name="projectnumber"
value="<?php ?>" readonly/>-->
<?php echo $options4;?>
But it would look like how you had it but instead of '1' inside the
box it would display '19000' and there would be nothing outside of the
box other than the label "Project Number". As far as i'm aware you can
assign a value to the text box, regardless of whatever the input is. I
would like it to display the value from one field name but actually
contain the value from a different field name. Both are in the same
table of course.
OK - gotcha. Unfortunately, you cannot do that. A textbox can only have one value and the user is always free to change that value, even if you make it read-only. You can test that out by using the developer toolbar in your browser. Probably a good time to mention that all user input should be considered dangerous and you should never trust it. Once they have submitted the form you need to verify it.
What I would recommend in your case is to use a hidden <input> which contains the value you actually want to submit; projectnoid. You can then display the Project Number in any manner you choose.
<form>
<h1>Project Number: 19000</h1>
<input type="hidden" name="projectnoid" value="1">
<input type="submit" name="submit">
</form>
To generate this, you would:
<?php
while($row4 = mysqli_fetch_row($result4)){
$projectnoid = $row[0];
$projectNumber = $row[1];
echo '<h1>' . $projectNumber . '</h1>';
echo '<input type="hidden" name="projectnoid" value="'. $projectnoid .'">
}
PHP:
<?php
$query_5 = "SELECT MAX(ProjectNumber) FROM tblProjects;";
$result_5 = mysqli_query($conn, $query_5);
$row_5 = mysqli_fetch_array($result_5);
$nextproject=$row_5['MAX(ProjectNumber)']+1;
?>
HTML:
<html>
<form class="myform" action="<?php echo htmlspecialchars($_SERVER[" PHP_SELF "]);?>" method="post">
<label for="txtfield">Project Number</label>
<input type="text" id="reqtxtfield" name="projectnumber" value="<?php echo $nextproject ?>" readonly/><br>
After running successful insert query:
echo "<meta http-equiv='refresh' content='0'>"; //REFRESH PAGE TO UPDATE PROJECT NUMBER
I have a created an HTML form where users sign up and input there data into an SQL database. I have then retrieved that data in a webpage where they can view their profile. I have created a page where users can edit there profile by creating a form which updates the value in the SQL database for there user id.
I would like the form to use the current value of the SQL cell as the default for that user to make alterations easier. Example: currently user 7 has their city set as New York, when they visits the edit info page, the city field in the form already hase New York as the default value.
I have no problem getting the SQL info and assigning it to a variable, I just don't understand how to set it as the default value. I am aware of how you set default values for input fields though.
My code:
<?php
$id = $_SESSION["user_id"];
// Create a query for the database
$query = "SELECT full_name FROM users WHERE id = $id LIMIT 1";
// Get a response from the database by sending the connection
// and the query
$response = #mysqli_query($dbc, $query);
// If the query executed properly proceed
if($response){
while($row = mysqli_fetch_array($response)){
echo $row['full_name'];
echo mysqli_error();
}
}
?>
<input type="text" name="aboutme" defualt="<?php echo $row['aboutme'] ?>" >
There's no default value for html input.
Input can has value, using attribute value:
<input type="text" name="some_name" value="Some value" />
In your case it's
<input type="text" name="aboutme" value="<?php echo $row['aboutme']?> />
Input can also has placeholder - some value that is present in an input, but erased when user starts to edit input's content:
<input type="text" name="aboutme" value="<?php echo $row['aboutme']?> placeholder="some value" />
How about
<?php
$id = $_SESSION["user_id"];
// Create a query for the database
$query = "SELECT full_name FROM users WHERE id = $id LIMIT 1";
// Get a response from the database by sending the connection
// and the query
$response = #mysqli_query($dbc, $query);
// If the query executed properly proceed
if($response){
while($row = mysqli_fetch_array($response)){
echo $row['full_name'];
?>
<input type="text" name="aboutme" value="<?php echo $row['aboutme'] ?>" >
<?php
echo mysqli_error();
}
}
?>
And here is a good example http://www.w3schools.com/php/showphpfile.asp?filename=demo_db_select_pdo
Neither of the answers worked and upon further research and trial and error I created a solution.
I changed the value that was store in the array to just be a normal php variable:
$aboutme = $row['aboutme'];
I then called that variable using the following code:
<input type="text" name="aboutme" value="<?php echo htmlspecialchars($aboutme); ?>" >
Thanks for your help.
I hope you find my answer useful.
Why don't you try using it as a place holder? This will provide editable text.
<input type="text" name="aboutme" placeholder="<?php echo $row['aboutme'];" />
As the title reveals I got an issue with how to update a checkbox that already has data in my SQL database.
My code looks like following:
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
Name <input type"text" name="inputName" value="<?php echo $hemsida['Namn']; ?>" /> </br>
Commentar <input type"text" name="inputComment" value="<?php echo $hemsida['Comment']; ?>" />
<br/>
</br><input type="checkbox" name="inputAll" value="checked" <?php echo $hemsida['All']; ?>/>Alla
<input type="hidden" name="id" value="<?php echo $_GET['id']; ?>" />
<input type="submit" name="submit" value="Redigera">
<?php
if(isset($_POST['submit'])) {
$u = "UPDATE hemsida SET `Namn`='$_POST[inputName]', `Comment`='$_POST[inputComment]', `ALL`='$_POST[inputALL]' WHERE ID = $_POST[id]";
mysql_query($u) or die(mysql_error());
echo "User has been modified";
header("Location: ..//sokh.php");
}
?>
The echo $hemsida['Namn'],['Comment'], and ['All'] just brings up and shows the old data thats in the database, but I do not understand what to do to update the checkbox. I have looked everywhere but I am stuck. Thank you in advance!
If I understand your question correctly, you are looking for a way to have a checkbox be either checked or not checked depending on database info. If so, I would try something like this. At the top of your code where you get your database info, put
if($conditionForCheck){
$inputAll = ' checked="checked"';
}
Then in your form
<input type="checkbox" name="inputAll"<?php echo $inputAll; ?> />
your question is not clear but i think you have a column in your database named "all" ? and perhaps this column can take only 1 value (true or false) !!
then you can test this value in your form, if the value is true : checkbox will be checked, else : checkbox will not be checked :
<input type="checkbox" name="inputAll" checked="<?php if($hemsida['All'] == true) echo checked; ?>" />Alla
dont use value="", use checked instead, then test value of $hemsida['All'] if it's true echo checked else anything to do
for your php code and server side of your application you can just test if checkbox is checked and then you have choice for what do you want to assign to your column in database, for example if checkbox is checked create a variable (for example $value_of_checkbox) and assign a value ("true" for exampel) to this variable, then include this variable in your sql code for update database column :
if (isset($_POST['inputALL'])) {
$value_of_checkbox = true;
}
else {
$value_of_checkbox = false;
}
if(isset($_POST['submit'])) {
$u = "UPDATE hemsida SET `Namn`='$_POST[inputName]', `Comment`='$_POST[inputComment]', `ALL`='$value_of_checkbox' WHERE ID = $_POST[id]";
mysql_query($u) or die(mysql_error());
echo "User has been modified";
header("Location: ..//sokh.php")
note : i change also sql code in this part : ALL='$value_of_checkbox'
I have a small (42 hours) problem with my code trying to edit article
- just the basic editNews.php
When I choose article to edit the data appears in the forms from the DB and when
I hit "update" it returns no error but the data wasn´t updated
<?PHP
connection to database blah blah
?>
<?php
if(isset($_POST['update']))
{
$newsid = $_POST['newsid'];
$date=$_POST['date'];
$time=$_POST['time'];
$location=$_POST['location'];
$result=mysql_query("UPDATE news SET date='$date',time='$time',location='$location', WHERE newsid=$newsid");
header("Location: listNews.php");
}
}
?>
<?php
$newsid = $_GET['newsid'];
$result=mysql_query("select * from news where newsid=$newsid");
while($res=mysql_fetch_array($result))
{
$date = $res['date'];
$time = $res['time'];
$location = $res['location'];
}
?>
This is the form - just the normal one....
<form method="post" action="editNews.php" name="form1">
each item is like
<input type="text" name="headline" value="<?php echo $location;?>" id="UserName">
and
<input type="hidden" name="newsid" value=<?php echo $_GET['newsid'];?>
<input name="update" type="submit" value="update" />
Most likely there is something that I don´t see but "seeing" has taken almost 2 days now
... Is there a possibility I don´t have "edit" privileges in the mySql?
How do you know there was no error? Your code lacks:
print mysql_error();
Add it right after the UPDATE query.
Also your code is most likely to fail whenever the submitted content itself contains single quotes. To send correct SQL to the database it's advisable to apply mysql_real_escape_string() on all input variables.
Try
$result= mysql_query('UPDATE news SET
date = "'. $date .'",
time = "'. $time. '",
location = "' .$location. '"
WHERE newsid = '.$newsid.';') OR die(mysql_error());
I'll try to keep it simple... this is the code I am using to populate a dropdown menu from a database. This populates the dropdown menu correctly.
<form action="formsubmit.php?team_id=6" method="post">
<label> <br />What did they say?: <br />
<textarea name="quotetext" rows="10" cols="26"></textarea></label>
<select name='name'>
<?php
while ($temp = mysql_fetch_assoc($query)) {
echo "<option value=" . $temp['id'] . ">".htmlspecialchars($temp['lastname']) . ", " . htmlspecialchars($temp['firstname']) . "</option>";
}
?>
</select>
<input type="submit" value="Submit!" />
</form>
What I'm attempting to do is pass the person's ID into along to the file formsubmit.php which is called on the submission of the form. When I use $die($sql) on my database query in formsubmit.php, the ID of the person is blank.... everything else gets passed through just fine. Here is the relevant code in formsubmit.php:
$quotetext = $_POST['quotetext'];
$id = $_POST['id'];
$team_id = $_GET['team_id'];
$sql = "INSERT INTO quotes SET
speaker_id='$id',
quotetext='$quotetext',
game_id=2";
die($sql);
EDIT: Fixed, credit to Michael.
<select name='name'>
Should be:
<select name='id'>
At first glance, it looks like a typo. Try:
$id = mysql_real_escape_string($_POST['name']);
In your form, you are using <select name='name'>, but in your PHP script you're calling for $_POST['id'].
Change it to:
<select name='id'>
The problem is that you don't have a form variable with "id" as its name. It looks to me as you just need to change <select name='name'> to <select name='id'>.
$team_id = $_POST['name']
Your select element is named "name" and your form is "POST".
You need to change
$id = $_POST['id'];
to
$id = $_POST['name']; // as name is what you gave your select element