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[SOLVED] Drop down Problem


boney alex

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Hi, I making an 'Add New Vehicle' feature using drop down lists for some of the data entry. If there is a probelm with the data entry such as a null field then the form is shown with all the previous entered data. How do I show what option has already been selected in a drop down list?

 

newvan.php

<form name="newvan" action="addnewvehicle.php" method="post">
<fieldset>
<legend>Vehicle Information</legend>
	<label for="registration">Registration:</label>
		<input type="text" input id="registration" name="registration" size="10"><br><br>
<?php
$result = mysql_query("SELECT MakeModelID, Make, Model, Specification FROM MakeandModel ORDER BY Make, Model, Specification");
?>
	<label for="branch">Type of Van:</label>
		<select name="vantype" STYLE="width: 220px" size="0">
		<option selected value=""></option> 
<?php 
while ($row = mysql_fetch_array($result)) {
echo "<option value=\"{$row[MakeModelID]}\">{$row[Make]} {$row[Model]} {$row[specification]}</option>\n";
} 
?>
		</select><br><br>			
<?php
$result = mysql_query("SELECT BranchID, Branch_Name FROM Branch ORDER BY Branch_Name");
?>
	<label for="branch">Branch:</label>
		<select name="branch" STYLE="width: 220px" size="0">
		<option selected value=""></option> 
<?php 
while ($row = mysql_fetch_array($result)) {
echo "<option value=\"{$row[branchID]}\">{$row[branch_Name]}</option>\n";
} 
?>
		</select><br><br>
	<label for="colour">Colour:</label>
		<input type="text" input id="colour" name="colour" size="20"><br><br>
	<label for="mileage">Mileage (e.g. 55000):</label>
		<input type="text" input id="mileage" name="mileage" size="10"><br><br>
	<label for="transmission">Transmission:</label>
		<select name="transmission" STYLE="width: 90px" size="0">
		<option selected value=""></option>
		<option value="Manual">Manual</option>
		<option value="Auto">Automatic</option>
		</select><br>												
</fieldset>

<br>
<center><input type="submit" value= "Add Vehicle" class="button" name="submit"></center>
</form>

 

I can show the normal text fields by POSTING the value into a variable and echo in the value:

 

 <label for="registration">Registration:</label>
<input type="text" input id="registration" name="registration" size="10" value="<? echo $registration; ?>"><br><br>
<label for="colour">Colour:</label>
<input type="text" input id="colour" name="colour" size="20" value="<? echo $colour; ?>"><br><br>
<label for="mileage">Mileage (e.g. 55000):</label>
<input type="text" input id="mileage" name="mileage" size="10" value="<? echo $mileage; ?>"><br><br>

 

Thanks for any help

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Hey, not sure if this is what you mean (you want to keep the same selections between pages?)

 

The below works for me...

 

<form action="/greetingscards.php" method="post" name = "themes" id="themes">
				<table width="90%">
					<tr width="100%">Theme</tr>
					<tr>
						<td width="50px"><select name="theme" id="theme1">
							<option value="all"<?php if($_POST['theme'] == "all") { echo ' selected="selected"'; } ?>>All</option>
							<option value="animal"<?php if($_POST['theme'] == "animal") { echo ' selected="selected"'; } ?>>Animal</option>
							<option value="floral"<?php if($_POST['theme'] == "floral") { echo ' selected="selected"'; } ?>>Floral</option>
							<option value="leisure"<?php if($_POST['theme'] == "leisure") { echo ' selected="selected"'; } ?>>Leisure</option>
							<option value="travel"<?php if($_POST['theme'] == "travel") { echo ' selected="selected"'; } ?>>Travel</option>
						</select></td>
						<td><input type="submit" name="submit3" value="Find cards!" /></td>
					</tr>
				</table>
			</form>

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thanks mate, that works great for my HTML populated drop down, BUT how would you go about keepin selections if the drop down list was populated from a table in your MySQL database??????? For example:

<?php
$result = mysql_query("SELECT BranchID, Branch_Name FROM Branch ORDER BY Branch_Name");
?>
	<label for="branch">Branch:</label>
		<select name="branch" STYLE="width: 220px" size="0">
		<option selected value=""></option> 
<?php 
while ($row = mysql_fetch_array($result)) {
echo "<option value=\"{$row[branchID]}\">{$row[branch_Name]}</option>\n";
} 
?>

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If I am getting it right this is what you would do like the previous post has but you would compare the sql result to the value in your option...something like this...

 

<select>

<option value="TEST" <? If ($sql[result]=='TEST') { echo "selected"; }?>>TEST</option>

</select>

 

It is much cleaner if your not breaking in and out of php to html and just use an echo statement to generate the html...

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