In this code i am trying to fetch city names into the html dropdown. Kindly corrct me if i am wrong anywhere, it give me an error
<?php
$query = Run("select city_name from City");
echo "<select name="city-name" style="width: 210px;">";
while ($row = mssql_num_rows($query))
{
echo "<option>$row->city_name</option>";
}
echo "</select>";
?>
use mysqli_fetch_array($query) insted of mssql_num_rows($query)
try this one
echo "<select>";
while ($row =mysqli_fetch_array($query))
{
echo "<option value='".$row['city_name']."'>".$row['city_name']."</option>";
}
echo "</select>";
mssql_num_rows returns the number of rows in the result set, it doesn't iterate and return individual rows. Try using mssql_fetch_object instead.
If you start a string with " then you have to escape all occurrences of " inside your string or use '.
For example:
echo "<select name=\"city-name\" style=\"width: 210px;\">";
or
echo '<select name="city-name" style="width: 210px;">';
so you don't accidently close the string.
Also like the others pointed out, you have to use
mysqli_fetch_array($query).
Use this
$query = Run("select city_name from City");
echo "<select name='city-name' style='width: 210px;'>";
while ($row = mssql_fetch_object($query))
{
echo "<option>$row->city_name</option>";
}
echo "</select>";
Use This
$query = Run("select `city_name` from City");
echo "<select name='city-name' style='width: 210px;'>";
while ($row = mssql_num_rows($query))
{
echo "<option value='.$row->city_name.'>".$row->city_name."</option>";
}
echo "</select>";
Related
So I have a code
<?php
$showorder = "SELECT order_number FROM orders WHERE customer_number=522";
$orderesult = mysqli_query($con, $showorder);
$ord = mysqli_fetch_array($orderesult);
?>
in my database customer number 522 has 2 order numbers, when i tried to show the result, it only shows 1.
Here's my other code
echo "<table>";
echo "<th>Order Number</th><th>Order date</th>";
echo "<tr><td>";
echo $ord["order_number"];
echo "</td><td>";
echo $ord["order_date"];
echo "</td></tr>";
You just need to use while() here for getting all records, something like:
while($ord = mysqli_fetch_array($orderesult)){
//echo all value here
}
Also note that, if you want to print $ord["order_date"] than you must need to select column also in your query.
Otherwise, $ord will only contain order_number value.
Your SQL is missing the extra column.
Current SQL:
SELECT order_number FROM orders WHERE customer_number=522
Change to:
SELECT order_number, order_date FROM orders WHERE customer_number=522
Put mysqli_fetch_array($orderesult); in a while loop.
while($ord = mysqli_fetch_array($orderesult)) {
echo $ord["order_number"];
# code
}
Replace your code with the below code and then try again
<?php
$showorder = "SELECT order_number, order_date FROM orders WHERE customer_number=522";
$orderesult = mysqli_query($con, $showorder);
echo "<table>";
echo "<tr>";
echo "<th>Order Number</th><th>Order date</th>";
echo "</tr>";
while($ord = mysqli_fetch_array($orderesult)) {
echo "<tr>";
echo "<td>$ord['order_number']</td>";
echo "<td>$ord['order_date']</td>";
echo "</tr>";
}
echo "</table>";
?>
echo "<table>";
echo "<th>Order Number</th>";
while($ord = mysqli_fetch_array($orderesult)) {
echo "<tr><td>";
echo $ord["order_number"];
echo "</td></tr>";
}
you must use loop to show all result , and you can use echo one time .
while($ord = mysqli_fetch_array($orderesult)) {
echo "<table>
<th>Order Number</th><th>Order date</th>
<tr><td>".
$ord["order_number"]."
</td></tr>";
}
When I add my first record into the database table, it doesn't show on the page where the records are displayed. But when I add the second record and on, they are displayed on the page except the first record that I entered.
Here is my code:
$result = mysql_query("SELECT * FROM members ORDER BY player_role DESC", $db);
while ($row = mysql_fetch_array($result))
{
echo "<table>";
echo"<tr><th><B>Player Name</B><Th><B>Role</B></TR>";
while ($myrow = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>". $myrow['player_name']. "</td>";
echo "</td>";
echo "<td>" .$myrow['player_role']. "</td>";
echo "</tr>";
}
echo "</table>";
}
Can someone please tell me what is wrong?
It may be caused by nested while() loop. No need to use nested while(). Use one while() instead. Example:
$result = mysql_query("SELECT * FROM members ORDER BY player_role DESC", $db);
echo "<table>";
echo"<tr><th><B>Player Name</B></th><th><B>Role</B></th></tr>";
while ($row = mysql_fetch_array($result))
{
echo '<tr><td>'.$row['player_name'].'</td><td>'.$row['player_role'].'</td></tr>';
}
echo "</table>";
Player Name and Role should not be inside while() loop.
MOST IMPORTANT Do not use mysql, it is deprecated. Instead use mysqli
Your code (changed)
$result = mysqli_query($db,"SELECT * FROM members ORDER BY player_role DESC");
echo "<table>";
echo"<tr><th><B>Player Name</B></th><th><B>Role</B></th></tr>";
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
echo '<tr><td>'.$row['player_name'].'</td><td>'.$row['player_role'].'</td></tr>';
}
echo "</table>";
I have four drop-down lists that I would like to populate with values from an MSSQL table. All four lists should contain the same values. The query looks like this:
$data = $con->prepare("SELECT ID, Code FROM Table WHERE Code = :value ORDER BY Code");
$input = array('value'=>'value'); //'value' is hardcoded, not a variable
$data->execute($input);
And here is the code for my drop-downs:
<?php
echo "<select name=\"proj1[]\">";
while($row = $data->fetch(PDO::FETCH_BOTH))
{
echo "<option value='".$row['Code']."'>".$row['Code']."</option> ";
}
echo "</select>";
?>
This works fine for one drop-down. If I try to create another one (proj2[], proj3[], proj4[]) and apply the same query, however, the PHP page stops loading at that point and the second drop-down does not populate. The only way I've found around it is to copy the query and change the variables ($data becomes $data2 for proj2[], and so on). I'd really rather not have to write the same query four times. Is there a way around it?
$select = '';
while($row = $data->fetch(PDO::FETCH_BOTH))
{
$select .= "<option value='".$row['Code']."'>".$row['Code']."</option> ";
}
echo "<select name=\"proj1[]\">";
echo $select;
echo "</select>";
echo "<select name=\"proj2[]\">";
echo $select;
echo "</select>";
//etc...
Why not just put all of it in a veriable and then using it 4 times?
Somthing like this:
<?php
while($row = $data->fetch(PDO::FETCH_BOTH))
{
$options .= "<option value='".$row['Code']."'>".$row['Code']."</option> ";
}
for($i = 0; $i <= 4; $i++){
echo "<select name=\"proj1[]\">";
echo $options;
echo "</select>";
}
?>
Ok so here is my code:
$select_status = 0;
$select_status = "<select name='status'>\n";
$select_status .= "<option value=''>SELECT ONE</option>\n";
$sdataset = mysql_query("SELECT id, name FROM phponly_category") or die(mysql_error());
while($srow=mysql_fetch_assoc($sdataset)) {
echo implode(", ", $srow);
echo "<br />";
$select_status .= "<option value='".$srow['name']."'";
$select_status .= ">".$srow['name']."</option>\n";
} // end while loop
echo "out of the loop";
$select_status .= "</select>\n";
// now insert the <select> list control into the page
echo $select_status;
The code works fine until the last row when it breaks. I cannot get the echo $select_status printed.
I have tried to see what is going on with the SQL query results by printing each row but everything looks fine there. For some reason, at the last row the while loop breaks and even the code after while loop doesn't get executed.
Don't do the or die(mysql_error()) portion in the while test...do it before.
if($sdataset==false) {
die(mysql_error());
}
while($srow=mysql_fetch_array($sdataset)) {
$select_status .= "<option value='".$srow['name']."'".">".$srow['name']."</option>\n";
} // end while loop
The or die() on your while() loop will actually kill the script when you read the end of the result set. mysql_fetch will return false, triggering the or die().
While checking for errors is good on queries, you can't do it like this on the fetching part, because you get false positives like this.
I personally don't like to echo out html code, if your goal is to do validate whether there's result comes out of the query, you can do something like this
<?php
$sdataset = mysql_query("SELECT id, name FROM phponly_category");
if (mysql_num_rows($sdataset) > 0) {
?>
<select name='status'>
<option value=''>SELECT ONE</option>
<?php
while($srow = mysql_fetch_array($sdataset)) {
?>
<option value='<?php echo $srow['name'] ?>'><?php echo $srow['name'] ?></option>
<?php } // end while loop ?>
</select>
<?php
} // end of if
else {
// Whatever you wanna put here
}
?>
EDITED: There's a typo at mysql_num_rows, try this one again
Use mysql_fetch_assoc so you can get the data using the field name as the key.
<?php
$conn = mysql_connect('localhost', 'username', 'password');
mysql_select_db('test');
$select_status = "<select name='status'>\n";
$select_status .= "<option value=''>SELECT ONE</option>\n";
$sdataset = mysql_query("SELECT id, name FROM phponly_category") or die(mysql_error());
while($srow=mysql_fetch_assoc($sdataset)) {
$select_status .= "<option value='".$srow['name']."'".">".$srow['name']."</option>\n";
} // end while loop
$select_status .= "</select>\n";
echo $select_status;
?>
Note: The use of this extension is discouraged. Take a look at mysqli or PDO.
$sql = "SELECT DISTINCT Branch FROM student_main";
$result = mysql_query($sql);
$row_num = mysql_num_rows($result);
$rows = mysql_fetch_array($result);
echo "<select name='Branch'>";
for($i=0;$i<=$row_num-1;$i++){
echo "<option value='".$rows[$i]."'>".$rows[$i]."</option>";
}
echo "</select>";
echo "<input type='submit' Value='submit' />";
echo "</form>";
I am trying to create a dropdown using the above code for my form. But its not working. There are 3 distinct values in the Branch column but in the dropdown, it shows only one value(the first one) and the next two as blank values.
However when in echo $row_num, its shows 3.
Thats means its fetching the three rows, but then why its not showing in the dropdown list.
If I run the same query in phpmyadmin it shows the correct answer i.r it returns 3 distinct Branch values.
You should do something like this:
$sql = "SELECT DISTINCT Branch FROM student_main";
$result = mysql_query($sql);
echo "<select name='Branch'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row[0]."'>".$row[0]."</option>";
}
echo "</select>";
echo "<input type='submit' Value='submit' />";
echo "</form>";
you need to mysql_fetch_array() for each row. That function returns an associative array for one row only. just include it inside your for loop just above your echo statement.
edit: mysql_fetch_array() actually returns an array (by default) that has associative indices and numbered indices. You can continue using it the same way, though.
You need to loop through your query using the following:
$sql = "SELECT DISTINCT Branch FROM student_main";
$result = mysql_query($sql);
echo "<select name='Branch'>";
while($rows = mysql_fetch_array($result)){ // should probably use mysql_fetch_assoc()
echo "<option value='".$rows['Branch']."'>".$rows['Branch']."</option>";
}
echo "</select>";
echo "<input type='submit' Value='submit' />";
echo "</form>";
mysql_fetch_array only returns the current dataset as an array, and moves the internal pointer ahead. You need to repeatedly call mysql_fetch_array to get all results.
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row['Branch']."'>".$row['Branch']."</option>";
}
There is a problem in the loop using a while loop:
while($rows=mysql_fetch_array($result)){
echo "<option value='".$rows[$i]."'>".$rows[$i]."</option>";
}
Try this
What you really need is to learn how to use templates.
But it seems Stackoverflow is definitely not the place where one can learn professional ways of website developing.
get your data first
$select = $array();
$sql = "SELECT DISTINCT Branch FROM student_main";
$res = mysql_query($sql) or trigger_error(mysql_error().$sql);
while($row = mysql_fetch_array($res)) $select = $row[];
And then use it in the template
<form>
<select name='Branch'>
<? foreach($select as $row): ?>
<option value="<?=htmlspecialchars($row['Branch'])?>">
<?=htmlspecialchars($row['Branch'])?>
</option>
<? endforeach ?>
</select>
<input type='submit' Value='submit' />
</form>
mysql_fetch_array will only return the first row...
see here for full details :)