I have below code which runs query correctly but does not keep selected option. I could get similar answers but not worked with this code. Here select options are in combination of user given and rest are from mysql column.
I am new to php, so any help will be appreciated.
<select name="tester">
<option value=""> -----------ALL----------- </option>
<?php
$dd_res=mysqli_query($conn, "Select DISTINCT tester from workflow1");
while($r=mysqli_fetch_row($dd_res))
{
echo "<option value='$r[0]'> $r[0] </option>";
}
?>
</select>
<input type="submit" name="find" value="find"/>
You can use the $_POST superglobal array to check the value of submitted form fields. In this case $_POST['tester'] contains the value of the select field after submitting the form.
echo "<option value='$r[0]'".($_POST['tester']==$r[0]?' selected':'')."> $r[0] </option>";
I am thinking you are asking to select the value passed into the form? If I am correct in that, you would do something like:
$selVal = $_REQUEST['tester'];
$val = $r[0];
echo '<option value="'.$val.'"';
if ($val == $selVal) {
echo ' selected="selected"';
}
echo '>'.$val.'</option>';
The $_REQUEST gets any POST, GET or COOKIE value and should be validated to make sure no hack attempts or anything. Once you have the value, you just add the selected attribute if it matches the value in the list.
Related
So got a small problem here, tried so many different solution but it always ends up sending a blank value to the database. So i got a dropdown with two values and one value that is never going to be allowed to make a POST. So if you accidentally click on the "Oppdater" button before selecting a value from the dropdown menu i dont want it to be send, but now it does and just adds a blank value to the database. Example:
<form action="" method="post">
<input type="hidden" name="id" value="' . $row['id'] . '">
<select name="status2" class="endre-status">
<option selected="selected" disabled="disabled">Change status</option>
<option value="Done">Done</option>
<option value="Pending">Pending</option>
</select>
<input type="submit" name="insert2" value="Oppdater">
</form>
Here you can see that i have it selected and disabled and that should mean it will not be sent when i check with isset(), but it still does and puts a blank value in my database.
Here is the code i use to update the database with the new value from the dropdown.
if (isset($_POST['insert2']))
{
$status2 = $_POST['status2'];
$id = $_POST['id'];
$sql2 = ("UPDATE test3 SET status='$status2' WHERE id='$id'");
if (mysqli_query($conn, $sql2)) {
header("Location: index.php");
exit;
} else {
echo "Error: " . $sql2 . "<br>" . mysqli_error($conn);
}}
Here it how it looks:
I have also tried with setting the value of the disabled option to "0" and checking with empty(), but it still sends "0" to the database. If someone can help me that would be very nice. Thank you.
If your options are static you can check it that if value is from your options or not?
in_array function can help you
for example:
$options = array("Done","Pending");
if (in_array($_POST['status2'], $options)) {
in line :
<option selected="selected" disabled="disabled">Change status</option>
you have mentioned selected so by default , the first option would be returned as selected option and it seems logical that it returns 0 .
Suggested solution :
set a break point on insert2 button and leave dropdown empty , then check the returned value . at last try to prevent code side to send that value by control statements such as If
I use the following code to select all value from a DB table and put it in dropdown. Here is the code :
<select name="dis_name">
<?php
$qu="Select DISTINCT name from root";
$res=mysqli_query($con,$qu);
while($r=mysqli_fetch_row($res))
{
echo "<option value='$r[0]'> $r[0] </option>";
}
?>
Based on previous code a value is selected and submitted. I want to show previously selected value in dropdown on another page so i use following code:
<select name="dis_name">
<option value="RAM"<?php echo $r20 == "RAM" ? " selected" : ""; ?>>RAM</option>
<option value="ROHAN"<?php echo $r20 == "ROHAN" ? " selected" : ""; ?>>ROHAN</option>
</select>
This code work fine till no new value added in DB table. But whenever i add new value i have to update edit code manually. Is there any way, so that dropdown previous selection updated automatically in edit page.
Add the condition in your loop, eg:
<?php
while($r=mysqli_fetch_row($res))
{
echo "<option value='$r[0]'";
if ($r[0] == $r20) echo " selected";
echo "> $r[0] </option>";
}
?>
If I get what you want (assuming that $r20 is the previously selected value:
<?php
$qu="Select DISTINCT name from root";
$res=mysqli_query($con,$qu);
while($r=mysqli_fetch_row($res))
{
echo "<option value='$r[0]'";
if($r[0]==$r20){echo 'selected="selected"';}
echo "> $r[0] </option>";
}
?>
this way the select will display as selected the option that match the $r20 value dinamically.
BTW, you should use id in your code. This way you could also have more than one user with the same name (but different ID). With DISTINCT, as you are doing, if you have more than one user with the same name you will get only one.
I can't seem to get this right. I am trying to make a select box that gets it values from the database sticky please see my code below :
<select name="cob_bank">
<?php while($row = mysqli_fetch_array($query,MYSQLI_ASSOC)){
if($row['Id']==$myId) {
echo '<option value="'.$row['Id'].'"selected="selected">'.$row['Name'].'</option>';
} else {
echo '<option value="'.$row['Id'].'">'.$row['Name'].'</option>';
}
}?>
</select>
after the select box is filled in and the submit button is pressed where I save my Post variable
$myId = $_POST['cob_name'];
The above code does not work I please assist
the no space could be causing the problem as a space before selected
if($row['Id']==$myId) {
echo '<option value="'.$row['Id'].'" selected="selected">'.$row['Name'].'</option>';
}
The select name should be set as:
<select name='cob_name'>
<select name="parentgroup" type="text" id="select2" onchange="this.form.submit();">
<?php
echo "<option value=\"0\">All Users</option>";
$result = mysql_query("select * from login WHERE stato=1");
while ($myresults = mysql_fetch_array($result))
{
$username = $myresults['user_name'];
echo "<option value=\"".$username."\" ";
echo $username== $parentgroupid ? " selected" : "";
echo ">".$username."</option>";
}
?>
</select>
Hi...am supposed to use the first element in <select> which is <option value=\"0\">All Users</option> to fetch values from a mysql database.
I want to now how you can assign this to a variable and use it just like $parentgroupid
When you submit the form (which presumably exists since you are referencing it from your JavaScript): Get the value from $_GET['name_of_form_control'] in the document referenced by the action attribute of the form.
When you submit the form, depending on the method specified in the form ("GET" or "POST"), the value will be in either the $_GET or $_POST arrays on the page to which you are submitting. You can retrieve the value by name (the name of the control):
$parentgroupid = $_GET['parentgroup'];
or
$parentgroupid = $_POST['parentgroup'];
I generated a drop down menu with the code below. How can I set the field to the GET variable after submit is clicked?
<select>
<?php
$sql="SELECT c FROM problems WHERE b='$id'";
$result=mysql_query($sql);
$options="";
while ($row=mysql_fetch_array($result)) {
$id=$row["c"];
$c=$row["c"];
$options.="<option value=\"$id\">".$c;
}
?>
<option value=''>C <?=$options?> </option>
</select>
<select>
<?php
$sql="SELECT c FROM problems WHERE b='$id'";
$result=mysql_query($sql);
$options="";
while ($row=mysql_fetch_array($result)) {
$id=$row["c"];
$c=$row["c"];
if($_GET['var'] == $id)
echo '<option selected="selected" value=\"$id\">' . $c . '</option>';
else
echo '<option value=\"$id\">' . $c . '</option>';
}
?>
</select>
Basic idea is compare value GET data with database data and using if else condition add selected="selected" if condition matched. I am directly printing string as they will not be getting use later on.
I'm a bit confused about what you want to know. To have an option selected by default, use the selected="selected" attribute on that option.
If you want to know how to get the submitted value in PHP, you need to assign the name attribute to the <select> tag. In general, however, you've got the idea right.