<?php
$id = $_SESSION['user_id'] ;
echo "<form method='post' action='#'>";
echo "</select>
<p>Which Hospital Would You Like to Submit To?</p>";
$queryitem = "SELECT * FROM vendor_hospital WHERE vendor_hospital.user_id = '$id' AND vendor_hospital.approval_status = '1'" or die('MYSQL error: ' . mysql_error());
if ($result = mysql_query($queryitem)) {
if ($success = mysql_num_rows($result) > 0) {
echo "<select name='hospital_name'>";
echo "<option>-- Select A Facility --</option>";
while ($row = mysql_fetch_array($result))
echo "<option value='$row[manufacturer_id]'>$row[manufacturer_id]</option>";
echo "</select><br><br>";
} else {
echo "No results found.";
}
} else {
echo "Failed to connect to database.";
}
echo "<input type='submit' value='Submit' name='submit' class='button' /></form>";
?>
For some reason I'm stuck here. I'm just trying to get the manufacturer name to show in my options instead of the manufacturer_id. The manufacturer name is a foreign key in another table so I can't simply call $row[manufacturer_id] in my option tag. What should I do here? My only thought is to run a query inside the option tag for every manufacturer_id listed as a value but I'm sure that is overkill. Can someone point me in the right direction of a more elegant solution than that?
Not really sure what you are trying to do...but try this.
Replace this:
while ($row = mysql_fetch_array($result))
echo "<option value='$row[manufacturer_id]'>$row[manufacturer_id]</option>";
echo "</select><br><br>";
With this:
while ($row = mysql_fetch_array($result)){
echo "<option value='".$row['manufacturer_id']."'>".$row['manufacturer_id']."</option>";
}
echo "</select><br><br>";
Also be sure to double check and make sure that you are pulling from the right database, that it is populated, you are calling the right table names, etc...
Related
<?php
$con = new mysqli('localhost', 'root' ,'', 'world');
$query = 'SELECT * FROM city ORDER BY Name';
if ($result = mysqli_query($con, $query)) {
echo "<table>";
while ($row = mysqli_fetch_assoc($result)) {
echo "<tr>"
echo "<td>" . $row['Name'] . "</td>";
echo "<td>" . $row['CountryCode'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_free_result($result);
}
mysqli_close($con);
?>
This simple code will display every entries in this database. Now I would like to add a selective display option by choosing the CountryCode.
$query2 = 'SELECT DISTINCT CountryCode FROM city ORDER BY CountryCode';
How do I use the result I got from the query above and make it become radio buttons to choose what to display?
Similar to what you are doing already: Something like
while ($row = mysqli_fetch_assoc($result)) {
echo "<input type='radio' name='whatever' value='".$row['Name']."'>". $row['Name'];
}
Ofcourse the field names can be whatever you like them to be, as long as you select them in the query
I'm having problems to find how to create an hyperlink in a table column result and then, on click, open another page with all fields (textboxes) filled. Imagine when a click an ID, i do a select * from table where column_id = ID... Is there a way to do it?
Thanks.
Best regards
I'm not completely sure what you are asking, but this may help you a bit.
First make a Javascript.
<script type="text/JavaScript">
function selectID() {
var ID = document.getElementById("ID").value;
document.location.href ="yoursite.php?ID="+ID;
}
</script>
Then connect to your database to query the table for a link ID (or more) for example by changing the variable $value.
<?php
//Connect to database
mysql_connect("host", "user", "pass");
mysql_select_db("db_name");
$value = 'something';
$ID = $_GET['ID'];
if (!$ID) {
$ID = 0;
}
if ($ID == 0) {
$query = "SELECT * FROM table WHERE `column_1` = '$value'";
$result = mysql_query($query);
echo "<table>";
while($myrow = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>";
echo "ID";
echo "</td>";
echo "</tr>";
}
echo "</table>";
}
elseif ($ID > 0) {
$query2 = "SELECT * FROM table WHERE `column_id` = '$ID'";
$result2 = mysql_query($query2);
while($myrow2 = mysql_fetch_array($result2)) {
$value1 = $myrow2['column_1'];
$value2 = $myrow2['column_2'];
}
echo "<form type=\"GET\" action=\"$PHP_SELF\">";
echo "<input type=\"text\" id=\"ID\" name=\"ID\" value=\"$ID\"><br>";
echo "<input type=\"text\" id=\"value1\" name=\"value1\" value=\"$value1\"><br>";
echo "<input type=\"text\" id=\"value2\" name=\"value2\" value=\"$value2\"><br>";
echo "<input type=\"hidden\" id=\"search\" name=\"search\" value=\"searching\">";
echo "<input type=\"submit\" id=\"submitbutton\" name=\"submitbutton\" value=\" Search \">";
echo "</form>";
}
?>
I have an error here, the data retrived from mysql database and listed in listbox or dropdown, when i need to update my form i get the result after refresh.
I have tried but need to sortout
<?
include("connect.php");
mysql_select_db("joblisting") or die(mysql_error());
$result = mysql_query("SELECT * FROM positiontitle") or die(mysql_error());
$ans=$positiontitle;
echo $ans;
?><select size=1 name="positiontitle" id="positiontitle"><?
echo '<option value="'.$ans.'">';
echo "</option>";
while ($row = mysql_fetch_array($result))
{
echo "<option>";
echo $row['positiontitle'];
echo "</option>";
}
echo "<br />";
?>
where are you defining $positiontitle ? shouldn't it be $_POST['positiontitle']?
I've been trying think of a way to do this. I want it to where users can check off items, hit submit and it goes to the code on the next page and deletes all of the checked items from the database. Problem one is that in the post its only sending over the last checked item. Here is how I have it set up right now.
echo "<form name='fm1' METHOD ='POST' ACTION ='displaydelete.php' > ";
//Draws up the table headers
echo "";
echo "";
echo "Fund Number ";
echo "Hours ";
echo "Percentage";
echo "Delete";
echo "";
//While there are query results data is pushed into table cells
while ($row = mysql_fetch_array($queryResult2))
{
$hours = $row['hours'];
$percentage = $hours / 160 * 100;
echo "<tr>";
echo "<td>";
echo $row['funnumber'];
echo "</td>";
echo "<td>";
echo $hours;
echo "</td>";
echo "<td>";
echo $percentage ."%";
echo "</td>";
echo "<td>";
echo "<input type='checkbox' name='id' value='$row[id]'/>";
echo "</td>";
echo "</tr>";
}
//End of tabel
echo "</table>";
echo" ";
echo "";
What I would like to do is push all of the items into a variable and maybe delete them that way. I'm not really sure how you would handle multiple deletes. I'm doing my delete like this for something else if this helps any.
$query = "DELETE FROM users
WHERE ninenumber = '$ninenumber'";
$result = mysql_query($query)
or die("Query Failed: " .mysql_error());
mysql_close($conn);
In your form:
<input type='checkbox' name='id[]' value='$row[id]'/>
Then, in the file you post to:
if(is_array($_POST['id'])){
foreach($_POST['id'] as $id){
...do something to $id;
}
}
Instead of this:
echo "<input type='checkbox' name='id' value='$row[id]'/>";
You need this:
echo "<input type='checkbox' name='id[]' value='$row[id]'/>";
Note the difference. I added [] after the input name. This tells the client and server that there are multiple inputs with that name. $_POST['id'] will be an array you can loop through on the next page.
foreach ($_POST['id'] as $checkbox) {
// DELETE FROM users WHERE ninenumber = $checkbox
}
isset, is_array, and mysql_real_escape_string omitted for brevity.
In the form-generating code, make the name in the html have" []" after it:
...
echo "<input type='checkbox' name='id[]' value='$row[id]'/>";
...
Then, in the form-reading code, your post'ed id will be an array.
$id_array = isset($_POST['id']) && is_array($_POST['id']) ? $_POST['id'] : array();
foreach( $id_array as $id ) {
$query = "DELETE FROM users WHERE ninenumber = '" . mysql_real_escape_string($id) . "'";
// execute the delete query
}
Putting [] after the name of a control will turn it into an array in the superglobal that you can then iterate over to get all the values from.
You need to have the same name for all of your checkboxes, then all values are passed as array POST variable.
<input type='checkbox' name='id[]' value='$row[id]'/>
Change
name=id
to
name=id[]
this will then give you an array.
What i'm trying to do is display a drop down with all field names from mysql database, once the user picks one and submits the form i want to display a second dropdown filled with all the rows from the submitted field name, this is my code so far:
$result = mysql_query("select * from `parts`") or die(mysql_error());
echo "<form action='".$_SERVER['PHP_SELF']."' method='post'>";
echo "<select name='field_names'>";
$i = 0;
while ($i < mysql_num_fields($result)) {
$fieldname = mysql_field_name($result, $i);
echo '<option value="'.$fieldname.'">'.$fieldname.'</option>';
$i++;
}
echo "</select>";
echo "<input type='submit' value='submit'></input>";
echo "</form>";
if($_POST) {
$fields = $_POST['field_names'];
$result1 = mysql_query("select '".$fields."' from `parts`") or die(mysql_error());
echo '<select name="fields">';
while ($row = mysql_fetch_array($result1)) {
echo "<option value=".$row[$fields].">".$row[$fields]."</option>";
}
echo '</select>';
}
Can anyone spot where i'm going wrong, thanks
there is a mistake on the line number 29
$result1 = mysql_query("select '" . $fields . "' from `parts`") or die(mysql_error());
you are using ' instead of `. Do as follows
$result1 = mysql_query("select `" . $fields . "` from `parts`") or die(mysql_error());
Hope your problem is solved.
As it stands now, the second set of selects will be issued OUTSIDE of your </form> tag, so will never get submitted with the rest of the form. At best, you should move the form closing tag to below the POST handler.
here database details
mysql_connect('hostname', 'username', 'password');
mysql_select_db('database-name');
$sql = "SELECT username FROM userregistraton";
$result = mysql_query($sql);
echo "<select name='username'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['username'] ."'>" . $row['username'] ."</option>";}
echo "</select>";
here username is the column of my table(userregistration)
it works perfectly