Display data from SQL database in a drop down menu
I have a database with names stored. I have my query of the database working, but lets say I have 5 names that I want to display in a drop down menu. How do I make the default text in the drop down menu display those 5 names?
Basically what I am trying to accomplish is this:
Query my database and store all the names of clients to a variable. Say there are 5 names in the database, I need those 5 names to be stored in a variable. And then for my d开发者_StackOverflow中文版rop down menu, normally I put the text in like this: < option>Single Floor< /option>
But how do I get those 5 names to appear in the drop down list?
Below is a simple pseudo-script that selects information from a database and outputs a select drop down box. You will need to replace *_fetch_array
with whatever DB Extension you are using, and $row['Value']
and $row['DisplayValue']
with the appropriate field names from your DB Schema.
<select name = 'iAmASelect' id = 'iAmaASelect'>
<?php
$DB_Rows = /* fetch data from database */;
while($row = *_fetch_array($DB_Rows))
echo("<option value = '" . $row['Value'] . "'>" . $row['DisplayValue'] . "</option>");
?>
</select>
The select will submit $row['Value']
to the form handler, while displaying $row['DisplayValue']
to the user in the drop down list.
If I understand it right you want a select box where the first option contains all the names AND each name also is an option.
Either implode your array with names into a string.
or use a for(each) loop.
$string ='';
foreach($rows as $k=>$names ){
$string.=$names.' ';
}
$string =trim($string).
HTML :
<select>
<option value='0' selected="selected"> <?=$string;?> </option>
## loop your names as options.
</select>
Just using the following code :
<select name="shipDec" id="shipDec">
<?php
for($i=0;$i<=200;$i++)
{
echo "<option value='".$i."'";
if($shipDec==$i)
{
echo 'selected="selected"';
}
echo ">".$i."</option>";
}
?>
</select>
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