I would like to start out saying I don't want to use Javascript....
I have an array which is returning correctly I believe, however I am only having the first entry in the array display in the drop down list.
Is this a problem with the array? or with this function?
foreach($array as $key=>$value){
$html = "<option value='$key'>$value</key>";
}
echo "<select name="process">$html</select>";
You have to use the concatenation operator (.):
$html = '';
foreach($array as $key => $value)
{
$html.= "<option value='$key'>$value</option>";
}
echo "<select name=\"process\">$html</select>";
However, looking at the function you posted earlier, mysql_fetch_assoc only returns a single row at a time. You need to loop over that, instead. The following should suffice:
function tasks_list($p_id) {
$project_tasks = array();
$p_id = (int)
$p_id;
$func_num_args = func_num_args();
$func_get_args = func_get_args();
$result = mysql_query("SELECT task_name FROM tasks WHERE project_id = $p_id");
while($project_tasks = mysql_fetch_assoc($result))
{
$html.= "<option value='".$project_tasks['task_name']."'>".$project_tasks['task_name']."</option>";
}
echo "<select name=\"process\">$html</select>";
}
You need to concat the strings. Start by initializing $html to an empty string before starting your for loop.
$html = '';
foreach($array as $key => $value)
{
$html .= "<option value='$key'>$value</key>";
// same as $html = $html . "<option value='$key'>$value</key>";
}
Note: There is also a typo in your code:
echo "<select name="process">$html</select>";
should either be:
echo "<select name=\"process\">$html</select>";
OR
echo "<select name='process'>$html</select>";
if that's what you meant.
$html = "";
foreach ($array as $key => $value){
$html .= "<option value='$key'>$value</option>";
}
echo "<select name='process'>$html</select>";
= reassigns; .= appends
If you use = it reassigns $html in every iteration, so that $html will contain the result of the last iteration - one single option.
$result = mysql_query("SELECT task_name FROM tasks WHERE project_id = $p_id"); while ($task = mysql_fetch_assoc($result)) $project_tasks[] = $task;