Dearest genius coders,
I have a form in which users create a Team from a number of selections. These selections are created from a drop down menu that pulls from a MySQL DB.
<select name="A1">
<?php
$queryA1 = "select p_Num, p_Name from players where p_Class = 'A'";
$resultA1 = mysql_query($queryA1);
while ($lineA1 = mysql_fetch_array($resultA1, MYSQL_ASSOC)) {
?>
<option value="<?php echo $lineA1['p_Num'];?>"> <?php echo $lineA1['p_Name'];?> </option>
<?php } ?>
</select>
This displays as shown. The player name (p_Name) in the list, but the INSERT takes the player number (p_Num).
From there, I have an insert created that takes the players and puts them in a Team database, but uses the p_Num (player number / key) to reference in the Player database.
I'm not having issues with the insert, what I need is to display the results after submission to the user, but I want to display the player name (p_Name) not the player number (p_Num).
How would I go about this?
-Nick
Related
I am currently developing an application which allows the user to pick an option from a dropdown, and list table data based on this option, and done so dynamically. I was pondering the thought of a search bar which allows the user to search for a country name (within a continent) and list said country name upon searching. The part which confuses me is can I make it so that the user can pick a continent from the dropdown, and input a country name into the search bar then click the same button and receive the listing?
I understand the importance of picking relevant code however I think it would be important for the reader to see both my home page, and action page to understand what I am trying to achieve.
Here is the index code:
require_once('repeat_code.php');
$db = dbConn();
//Make an SQL statement
$sqlContinents = "SELECT DISTINCT ID as contID, Name as contName from w_Continent order by contName;";
//Execute SQL statement
$stmt = $db->query($sqlContinents);
//Start a form
echo "<form action='listCont.php' method='get'>\n";
//Start a select box
echo "<select name='contID'>\n";
//Loop through all continents
while ($continent = $stmt->fetchObject()) {
//Display each one as an option in the dropdown
echo "\t\t<option value='{$continent->contID}'> {$continent->contName} </option>\n";
}//end loop
// end select box
echo "</select>\n";
// display submit button
echo "<input type='submit' value='Find Country' />\n";
// end form
echo "</form>\n";
Here is the action page:
if(!empty($contID)) {
//Connect to the database
require_once('repeat_code.php');
$db = dbConn();
//Create SQL statement using ID
$sqlCountries = "SELECT w_Country.Name, w_Continent.Name as 'contName', w_Country.Region as 'regionname', w_Country.HeadOfState, w_Country.Capital
FROM w_Continent JOIN w_Country on w_Continent.ID = w_Country.Continent
WHERE w_Continent.ID = '$contID'";
//Execute statement to get a record set back
$stmt = $db->query($sqlCountries);
// Start a table
echo "<table border='1'>\n";
// Start a header row
echo "<tr><th>Country</th><th>Continent</th>\n";
//Loop through the record set
while ($continent = $stmt->fetchObject()) {
//Display each student in a row
echo "\t<tr><td>{$continent->Name}</td><td>$continent->contName</td>\n";
}//End loop
//end the table
echo "</table>";
Any help is appreciated.
Create a drop down and input box in a single line.
Drop Down Contain the continent and input box is empty.
In database create 2 tables in following format.
Continent Table
id | Continent_name
Countries Table
id | country_name | continent_id
Now run query in first table and display all the continent in drop down.
now when user select the continent name from the drop down and also write country name in the search box, then you have to query like this.
<form method="post">
<input type="hidden" name="command" value="search">
<select name="select_box">
<option value ="1">Asia</option>
<option value ="2">Africa</option>
</select>
<input type ="text" name="search_box" placeholder="Type Country">
</form>
Now Run PHP code like below.
if($_REQUEST['command']=='search'){
$continent_id = $_REQUEST['select_box']; //suppose user select asia
$country_name = $_REQUEST['search_box']; // suppose user type "Pakistan"
//now execute below query in countries table.
$query ="select * from countries where continent_id='$continent_id' and country_name like '%$country_name%'";
}
I need a little kickstart here.
We're just learning php in school and we have this project where we're making a website for movie browsing. So we wanna be able to select genre, and show all movies from that genre using a MySQL database. We're all clear on the SQL queries and such, my question is rather how I make the browser show movies depending on SQL query?
Let me explain. Say we're movies.com
So on movies.com/genre is where you select the genre right, and on movies.com/display is where you're supposed to see the movies from the genre selected. So, clicking on "Comedy" should take you to movies.com/display and show you only the comedy movies. Selecting "Drama" should take you to the same site (movies.com/display) and show you only the drama movies.
Problem here is that we just don't know where to begin, it became a problem when switching page to show certain sql queries depending on what you selected in a previous page. I am not sure how to Google it, but just a link or a suggestion will help. I'm sure it can't really be too hard.
Thanks in advance.
Start with the html to choose a genre:
<form action="display.php" method="POST">
<select name="genre">
<option value="drama">Drama</option>
<option value="comedy">Comedy</option>
<option value="thriller">Thriller</option>
<option value="horror">Horror</option>
</select>
<input type="submit" value="Submit"/>
</form>
Here we have a form with a dropdown menu to select the genre. The forms action goes to display.php
Create display.php where we can get the submitted value and put that into our query:
// get the submitted value
$genre = $_POST['genre'];
// set a default genre if the POST genre is empty
if(empty($genre)) {
$genre = 'comedy';
}
// connect to database
$link = mysqli_connect("localhost", "my_user", "my_password", "db_name");
// check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: (" . mysqli_connect_errno() . ") ";
}
// build the query
$query = "SELECT * FROM movie WHERE genre = '".$genre."'"; // unsafe !!
$query = "SELECT * FROM movie WHERE genre = '".mysqli_real_escape_string($link, $genre)."'"; // safer
// execute query
$result = mysqli_query($link, $query);
// check result
if (!$result) {
echo 'Query unsuccessfull!';
}
// show values
while($row = mysqli_fetch_assoc($result)) {
echo $row['title'];
echo $row['description'];
echo "<br/>";
}
// close connection
mysqli_close($link);
I would suggest having a table of genres. That way you can display the list of genres easily for a user to select from (ie, you could populate an HTML forms select list from that table). The list of genres would have a unique id (an integer) and a text description, so when the list of produced you display the description but have a value that is the id.
Then have a table of movies (I presume you already have this). Each movie would be identified by a unique integer id field/
Lastly a table of movies to genres. This would just link the id of the movie to the id of the genre(s) for the movie. This way you can have a movie linked to several genres if required.
Doing it this way also means that the values you return from the HTML forms can just be integers, which are easily made safe (just use the php intval() function).
In use you would provide a list of genres as a SELECT list. When the user selects an item and submits the form the id of the selected genre is returned to the script. The script can then do a query that joins the tables together, checking the genre is the selected genre that has been returned to the script.
For example.
tbl_genres
id
genre_name
tbl_movies
id
movie_name
tbl_genre_movie
genre_id
movie_id
Then to get the details for a returned genre id:-
$sql = "SELECT *
FROM tbl_genres
INNER JOIN tbl_genre_movie
ON tbl_genres.id = tbl_genre_movie.genre_id
INNER JOIN tbl_movies
ON tbl_genre_movie.movie_id = tbl_movies.id
WHERE tbl_genres.id = ".intval($_POST['genre_id']);
I have a simple drop down menu.
<form method="post" action="index.php">
<select name="mountname">
<option value="white">white</option>
<option value="black">black</option>
<option value="yellow">yellow</option>
<option value="green">green</option>
</select>
<input type="submit" value="Submit Pick" />
to save what is selected I used.
if (!empty($_POST['color'])){
$id = $_SESSION['user_id'];
$color = $_POST['color'];
mysqli_query($mysqli,"UPDATE users SET home_color='".$color."' WHERE id='".$id."'")or die("error == ----> ".mysqli_error());
mysqli_close($mysqli);
header('Location: index.php');
}
saving the color to mysql is no problem.
//update//
the USER table is set up like this.
ID, username,password, first_name, last_name, email,home_color
When a user selects his home color, and then submits it it is saved to the db.
ie 1, Bob, MD5pass, Bob,Smith,Bob#bob.com, Black
2, Joe, MD5pass, Joe,Doe,joe#Doe.com, Green
now i have another table called mount.
mount has color info in it.
this table hold color name, and info.
ID, color_name, color_info
the ID is is an INT with A_I.
Bob Logs in and selects his home color saves it to his profile.
so now when a person goes to bobs profile the will see color info.
how do I make it where it reads profile info and displays info from another table.
something like the code below. I know the code is wrong, but only way i can explain it.
if (black){
mysqli_query($mysqli,"SELECT * FROM mount;
}else{
(green)
As you said, color is saved in user's table. Problem is how to fetch color's data from mount along with user's data..
On Profile Page, you can get details of user with color by ..
$id = $_SESSION['user_id'];
$res = mysqli_query($mysqli,"SELECT * FROM users u, mount c WHERE u.home_color=c.color_name AND u.id='".$id."'")or die("error == ----> ".mysqli_error());
$result = mysqli_fetch_array($res);
mysqli_close($mysqli);
Now, you can manipulate $result as it contains users detail as well as color detail! We are getting data from both tables users and mount by same keys comparision home_color of users and color of mount.
If you saved colors in mount like 'Black', I'm suggesting you to keep same keyword in <select> dropdown, as 'Black' is not equal to 'black'.
Another thing I want to suggest you is, Change your select dropdown to dynamic. Fetch color data from table mount and use it like..
<select name="mountname">
<?php
$res_colors = mysqli_query($mysqli, "SELECT * FROM mount");
$colors = mysqli_fetch_array($res_colors);
foreach($colors as $color){
?>
<option value="<?php echo $color['color']; ?>"><?php echo $color['color']; ?></option>
<?php } ?>
</select>
Best way to use primary key as foreign key, so use ID of mount instead of color for dropdown and saving it to user's table.
I am trying to write a script to display records from an SQL database, but based on three variables which are set by three dropdown boxes on the page which are auto-populated from the database.
This is for a Learning Management System I am working on to provide effective feedback from online learning tests.
The code I am currently working on is below, along with the pseudo code which I hope will explain my requirements. The problem I am having is I cannot get the dropdown boxes to populate based on the criteria I have set in my pseudo code.
Any help is very much appreciated,
Thank you
John
// I used this article for the structure of the following script:
http://forums.devarticles.com/mysql-development-50/drop-down-menu-populated-from-a-mysql-database-1811.html
// Dropdown Box 1 - Choose the course - Show entries from the column "Name" from table "mdl_scorm". Once an option has been selected set variable $coursechoice to the value in the "id" column of the "mdl_scorm" table
// Dropdown Box 2 - Choose the user - Show entries from the columns "firstname" + "lastname" from table "mdl_user" IF the number shown in the "id" column of table "mdl_user" is present in the "userid" column of table "mdl_scorm_scoes_track" AND IF $coursechoice is present in the "scormid" column of table "mdl_scorm_scoes_track". Once an option has been selected set variable $userchoice to the value in the "id" column of table "mdl_user"
// Dropdown Box 3 - Choose the attempt - Show entries from the column "attempt" from table "mdl_scorm_scoes_track" IF $coursechoice is present in the "scormid" column of the table "mdl_scorm_scoes_track" AND IF $userchoice is present in the "userid" column of table "mdl_scorm_scoes_track". Once an option has been selected set variable $attemptchoice to the value in the "attempt" column from table "mdl_scorm_scoes_track"
// Submit button displays the records from table "mdl_scorm_scoes_track" which have a value in the column "scormid" which matches $coursechoice AND have a value in the column "userid" which matches $userchoice AND have a value in the column "attempt" which matches $attemptchoice
$sql="SELECT name FROM mdl_scorm";
$result=mysql_query($sql);
$options="";
while ($row=mysql_fetch_array($result)) {
$id=$row["id"];
$thing=$row["name"];
$options.="<OPTION VALUE=\"$id\">".$thing;
}
?>
<SELECT NAME=course>
<OPTION VALUE=0>Choose the course
<?=$options?>
</SELECT>
<?php
$sql="SELECT username FROM mdl_user";
$result=mysql_query($sql);
$options="";
while ($row=mysql_fetch_array($result)) {
$id=$row["id"];
$thing=$row["name"];
$options.="<OPTION VALUE=\"$id\">".$thing;
}
?>
<SELECT NAME=user>
<OPTION VALUE=0>Choose the user
<?=$options?>
</SELECT>
<?php
$sql="SELECT attempt FROM mdl_scorm_scoes_track";
$result=mysql_query($sql);
$options="";
while ($row=mysql_fetch_array($result)) {
$id=$row["id"];
$thing=$row["name"];
$options.="<OPTION VALUE=\"$id\">".$thing;
}
?>
<SELECT NAME=attempt>
<OPTION VALUE=0>Choose the attempt
<?=$options?>
</SELECT>
<?php
$finalresult = SELECT element, value FROM mdl_scorm_scoes_track WHERE scormid=$coursechoice AND userid=$userchoice AND attempt=$attemptchoice
while ($testrows = mysqli_fetch_array($finalresult)){
echo $testrows['value'];
I think the recommendation below will get your dropdowns working... To actually pull information dynamically based on the dropdowns will require a form, a submit button and some parsing of the $_POST variable.
One step at a time, lets get your dropdowns working.
First, don't forget to close your options with </option> after .$thing:
$options.="<OPTION VALUE=\"$id\">".$thing."</OPTION>";
and
<OPTION VALUE=0>Choose the course</OPTION>
In addition, you are only selecting the name, if you want the id you need to select that too.
$sql="SELECT name, id FROM mdl_scorm";
and finally, to use the results of the query as an associative array you need mysql_fetch_assoc
while ($row=mysql_fetch_assoc($result)) {
That last was was probably what was tripping you up most.
Hope that helps.
I have some trouble with this.
I have one database with the following tables:
Countries -> All countries of the world are added
Cities -> The cities are also added
user_profile -> The profile of the user with the fields "country" & "city".
Everything works fine, even the populating of the boxes is working. But I don't know how to get the SELECTED value of the user for both country & city.
I have the following code:
Select Country: <br />
<select id="countries">
<option value="">--</option>
<?php
$sql = "SELECT country, title FROM countries ORDER BY title ASC";
$rs = mysql_query($sql);
while($row = mysql_fetch_array($rs))
{
echo "<option value=\"".$row['country']."\">".$row['title']."\n ";
}
?>
</select>
Select City: <br />
<select id="cities">
<option value="">Select one</option>
<?php
$sql = "SELECT country, title FROM cities ".
"ORDER BY title ASC";
$rs = mysql_query($sql);
while($row = mysql_fetch_array($rs))
{
echo "<option class=\"".$row['country']."\" value=\"".$row['title']."\">".$row['title']."\n ";
}
?>
</select>
Now I need to check in the table user_profile which country & city he chose and show it to him.
I also use jchained (jquery plugin) to make both select boxes working together.
How can I achieve this? If I need to post anything else, please let me know.
Thanks in advance.
You have a lot of missing code here but I will try to help.
I am going to start by assuming that you have already stored the users selections in the user_profile table and that they are foreign keys to to correct relations.
To get the selected id's for repopulating the select boxes with the selected flag use:
SELECT country, city FROM user_profile where id = $user_id;
To get the text values you would do something like:
SELECT country.title as country, city.title FROM user_profile LEFT JOIN countries ON user_profile.country = country.id LEFT JOIN cities ON user_profile.city = cities.id WHERE user_profile.id = $user_id;
If you are not storing them as foreign key relations but instead by string title in the user_profile table you will need to do this:
SELECT country.title as country, city.title FROM user_profile LEFT JOIN countries ON user_profile.country = country.title LEFT JOIN cities ON user_profile.city = cities.title WHERE user_profile.id = $user_id;
You can then use these results to either display to the user or set the "SELECTED" flag in the options box.
Do you just mean you don't know what to do after they select the values? If so then you would just use the form to post the values to the new page. On the receiving page you can run a query to Update the user's profile.
Sorry if my response isn't detailed enough ><
edit Oh, I may see what you mean now. You're able to store it correctly but you want the correct option to be flagged as selected if it's what the user has previously set? If that's it then you can query for the value, then inside of the while loop just add a check to see if $user_country == $row['country'] and if so echo ' SELECTED '. Hope that helps!