Just like the title says i'm having difficulties in achieving it.
Here's my dropdownlist:
<?php
$query = "SELECT data, rel_id FROM $tbl_rel_balansas INNER JOIN $tbl_balansas ON $tbl_rel_balansas.rel_id = $tbl_balansas.id WHERE $tbl_rel_balansas.member_id = '$_SESSION[id]' group by data";
$result = mysql_query ($query);
echo "<select name=data value=''>Data</option>";
while($nt=mysql_fetch_array($result)){
echo "<option value=$nt[data] name=\"blabla\">$nt[data]</option>";
}
echo "</select>";
?>
Here's the buttonclick:
<?php
if(isset($_POST['Submit']))
{
$query = "SELECT SUM(suma), paskirtis FROM $tbl_rel_balansas INNER JOIN $tbl_balansas ON $tbl_rel_balansas.rel_id = $tbl_balansas.id WHERE $tbl_rel_balansas.member_id = '$_SESSION[id]' AND data ='".$_POST['data']."' group by paskirtis";
$result = mysql_query ($query);
echo "<tr><td>Paskirtis:</td><td>Biudzetas:</td><td>Isleista:</td><td>Likutis:</td></tr>";
while($nt=mysql_fetch_array($result)){
if($nt['SUM(suma)'] != null){
$suma = $nt['SUM(suma)'];
}
echo "<tr><td>$nt[paskirtis]</td>
<td><input type=\"text\" name=\"isleista[]\" value=\"Skiriamų pinigų kiekis...\" method=\"post\"></td><td>".$suma." Lt</td><td>--</td></tr> <br>";
}
}
?>
After I press it, it retrieves the data I want from the date I've chosen from the drop down list and also reset whole drop down list showing the first value of the dates from sql database, not the one I selected. If anyone knows how to keep the selected value in the list any help is greatly appriciated!
Try this, you need to place select="selected" in the while loop. See below code how I placed the $selected
<?php
$query = "SELECT data, rel_id FROM $tbl_rel_balansas INNER JOIN $tbl_balansas ON $tbl_rel_balansas.rel_id = $tbl_balansas.id WHERE $tbl_rel_balansas.member_id = '$_SESSION[id]' group by data";
$result = mysql_query ($query);
echo "<select name=data value=''>Data</option>";
while($nt=mysql_fetch_array($result)){
$selected = ($_POST['blabla'] == $nt[data])?'selected="selected"':NULL;
echo "<option value=$nt[data] name=\"blabla\" $selected >$nt[data]</option>";
}
echo "</select>";
?>
Related
EDIT: IGNORE ANY SQL INJECTIONS OR VULNERABLE CODE STATEMENTS :D
(School Project).
I wish to create a insert form on my webpage where I can select an artist from a table, including a song from a table and combine them for an insert into a combined foreign key table.
I have managed to do selects and insert with only individual artist and song drop-downs on my web-page, but would wish for combining the two ID's from each table to combine them to a many to many relative table. But when I press the submit button nothing happens, and I'm a beginner and don't know if I'm missing any important bits of actually Posting the information.
For troubleshooting I have tried my code, and tested it. I see if I remove my code theres no problem, so the problem persists on the syntax I believe, as the first dropdown shows, alongside the second dropdown and submit button, but the problem is within the actual processing and SQL query part, where it never goes to the DB..
The problem:
As you can see below I have a the text Song Name appear with a drop-down menu in the bottom left corner including the Artist Name with a submit button. But my problem persists as the select and then insert from the two drop downs into the combined table does not work, it does not actually submit, I want it to post into the DB what can I do. But somethings off? I would appreciate any questions or help, this community is so amazing and wonderful to operate in!
Database
PHP
<form method='POST'>
<?php
include('connect_mysql.php');
if(isset($_POST["mangetilmange"])) {
$song_id = $_POST["song_id"];
$artist_id = $_POST["artist_id"];
$sql ="INSERT INTO artist_has_song (song_id, artist_id) VALUES
('$song_id', '$artist_id')";
if($conn->query($sql)) {
echo "Completed";
} else {
echo "Blablalbablablablablablablabl $sql
($conn->error.";
}
}
?>
Song Name
<?php
$sql = "SELECT * FROM song";
$resultat = $conn->query($sql);
echo "<select name='song_id'>";
while ($rad = $resultat->fetch_assoc()) {
$song_id = $rad["song_id"];
$songname = $rad["songname"];
echo "<option value='$song_id'>$songname</option>";
}
echo "</select>";
?>
Artist Name
<?php
$sql = "SELECT * FROM artist";
$resultat = $conn->query($sql);
echo "<select name='artist_id'>";
while ($rad = $resultat->fetch_assoc()) {
$artist_id = $rad["artist_id"];
$artistname = $rad["artistname"];
echo "<option value='$artist_id'>$artistname</option>";
}
echo "</select>";
?>
</form>
<input type="submit" name="mangetilmange" value ="Submit">
change you code to this:
<form method='POST'>
<?php
include('connect_mysql.php');
if(isset($_POST["mangetilmange"])) {
$song_id = $_POST["song_id"];
$artist_id = $_POST["artist_id"];
$sql ="INSERT INTO artist_has_song (song_id, artist_id) VALUES
('$song_id', '$artist_id')";
if($conn->query($sql)) {
echo "Completed";
} else {
echo "Blablalbablablablablablablabl";
}
}
?>
Song Name
<?php
$sql = "SELECT * FROM song";
$resultat = $conn->query($sql);
echo "<select name='song_id'>";
while ($rad = $resultat->fetch_assoc()) {
$song_id = $rad["song_id"];
$songname = $rad["songname"];
echo "<option value='$song_id'>$songname</option>";
}
echo "</select>";
?>
Artist Name
<?php
$sql = "SELECT * FROM artist";
$resultat = $conn->query($sql);
echo "<select name='artist_id'>";
while ($rad = $resultat->fetch_assoc()) {
$artist_id = $rad["artist_id"];
$artistname = $rad["artistname"];
echo "<option value='$artist_id'>$artistname</option>";
}
echo "</select>";
?>
<input type="submit" name="mangetilmange" value ="Submit">
</form>
I'm having a hard time figuring this out..
I have two tables "teacher_info" and "section_info".
In my forms I included all the attributes in my section_info table except I used teacher name instead of teacher id for easy selection which is a dropdownlist of teacher's name, this is my code
<?php
include("anhsis.php");
mysqli_select_db($con,"anhsis");
$result= mysqli_query($con,"SELECT t_lname,t_fname FROM teacher_info");
echo"<select name='adviser' class='form-control' required>";
echo"<option value='0'>--Select Adviser--</option>";
while ($row=mysqli_fetch_array($result)) {
echo "<option value='".$row['t_lname']."".$row['t_fname']."'>".$row['t_lname'].", ".$row['t_fname']."</option>";
}
echo'</select>'
?>
and heres my php code to insert data in "section_info"
<?php
include_once('anhsis.php');
$room_id = $_POST['room_id'];
$section = $_POST['section'];
$adviser = $_POST['teacher_id'];
$level = $_POST['level'];
$curriculum = $_POST['curriculum'];
mysqli_select_db($con,"anhsis");
$result= mysqli_query($con,"SELECT * FROM section_info WHERE room_id= '$room_id'");
if (mysqli_num_rows($result)>0){
echo '<script type="text/javascript">';
echo 'alert("TIN No already exist!")';
echo '</script>';
}
else{
mysqli_query($con,"INSERT INTO section_info VALUES('$room_id','$section','$adviser','$level','$curriculum')");
}
?>
my problem is that in my section_info theres no attribute of teacher's name, instead it has teacher_id. So how am I going to insert teacher_id from the "teacher_info" table to the "section_info" table by just selecting the teacher's name in my dropdownlist. Or is it possible?
Thanks!
$result= mysqli_query($con,"SELECT t_lname,t_fname,teacher_id FROM teacher_info");
echo "<select name='adviser' class='form-control' required>";
echo "<option value='0'>--Select Adviser--</option>";
while ($row=mysqli_fetch_array($result)) {
echo "<option value='".$row['teacher_id']."'>".$row['t_lname'].", ".$row['t_fname']."</option>";
}
This way you will have teacher_id value in $adviser variable, ready to use.
I have a drop down that is being populated from a static select. Then when a choice is made in the first drop down a query runs and the second drop down is populated from the database depending on selection in first select box. Here is the code. I'm having a problem displaying the second drop down with the data.
$selectvalue = mysqli_real_escape_string($mysqli, $_GET['selectvalue']);
$result = mysqli_query($mysqli, "SELECT DISTINCT '$selectvalue' FROM accounts ");
echo '<option value="">Please select...</option>';
while($row = mysqli_fetch_array($result))
{
echo '<option value="'.$row['$selectvalue'].'">' . $row['$selectvalue'] . "</option>";
//echo $row['drink_type'] ."<br/>";
}
mysqli_free_result($result);
mysqli_close($connection);
?>
<?php
$selectvalue = mysqli_real_escape_string($mysqli, $_GET['selectvalue']);
$result = mysqli_query($mysqli, "SELECT DISTINCT * FROM accounts WHERE col_name = '".$selectvalue."' ");
echo '
<select name="some_name">
<option value="">Please select...</option>';
while($row = mysqli_fetch_array($result)) {
echo '<option value="'.$row['col_name'].'">'.$row['col_name']."</option>";
}
mysqli_free_result($result);
mysqli_close($connection);
?>
I am trying to embed a drop down list to an update page, it works fine but I am having problems with option selected part. When an option is shown it also duplicates it in the list.
Is there any way I can say if record is used do not use it again with PHP?
<?php
$sql = "SELECT TeamName, TeamID FROM tblTeam";
$result = mysql_query($sql);
$player_id = $_GET['id'];
$current_team = mysql_query("SELECT
tblteam.TeamID,
tblteam.TeamName,
tblplayer.PlayerID,
tblplayer.PlayerTeam,
tblplayer.PlayerName
FROM
tblplayer
INNER JOIN tblteam ON tblplayer.PlayerTeam = tblteam.TeamID
WHERE PlayerID = $player_id LIMIT 1 ");
$my_row = mysql_fetch_array($current_team);
?>
<select name="TeamName">
<option selected value="<?php echo $my_row['TeamID']; ?>"> <?php echo $my_row['TeamName']; ?> </option>
<?php
while ($row = mysql_fetch_array($result)) {
$team_name= $row["TeamName"];
$team_id = $row["TeamID"];
echo "<option value=\"$team_id\">$team_name</option>";
}
echo "</select>";
?>
It seems like simple if condition will solve the case for you:
while ($row = mysql_fetch_array($result)) {
$team_name= $row["TeamName"];
$team_id = $row["TeamID"];
if($team_id != $my_row['TeamID']){
echo "<option value=\"$team_id\">$team_name</option>";
}
}
Additionally you should always sanitize $_GET / $_POST params, in your example:
$player_id = intval($_GET['id']);
Intval will return 0 if the given format is not numeric, so your sql query is safe from this moment.
I have two SQL queries that display different results from my database. I'm using these results as navigation links in my nav bar.
At the moment all the results display as a single line of text, I want to make each SQL query display as a drop down menu, with all the results from the query as an option in the drop down.
Here's the code I'm using:
<?php
$q = "SELECT cat_id, cat_name FROM Category";
$result = mysqli_query($_SESSION['conn'], $q);
while ($row = mysqli_fetch_row($result)) {
echo "<a href='category.php?id=$row[0]'>$row[1]</a> ";
//display product categories
}
mysqli_free_result($result); //free result
$q = "SELECT brand_id, brand_name FROM Brand";
$result = mysqli_query($_SESSION['conn'], $q);
while ($row = mysqli_fetch_row($result)) {
echo "<a href='brand.php?id=$row[0]'>$row[1]</a> ";
//display product Brands
}
?>
use select option
echo "<select name='category'>";
while ($row = mysqli_fetch_row($result)){
echo "<option value='$row[0]'>$row[1]</option> ";
}
echo "</select>";
$q="SELECT cat_id, cat_name FROM Category";
$result = mysqli_query($_SESSION['conn'],$q);
$option1.="<select name='category'>";
while ($row = mysqli_fetch_row($result)){
$option1.="<option value='$row[0]'>$row[1]</option> ";
}
$option1.="</select>";
same for second
print this $option1 value in your view file
<select onchange="document.location.href='category.php?id='+this.value;">
<?php
$result = mysqli_query($_SESSION['conn'],$q);
while ($row = mysqli_fetch_row($result)):?>
<option value="<?=$row[0];?>"><?=$row[1];?></option>
<?php endwhile;?>
</select>