[英]Select option with info from database
I'm trying to show up in a select, multiple options that are in the table on database. 我试图显示在数据库表中的一个选择的多个选项中。
$query_lista = sqlsrv_query($conn, "SELECT * FROM inquerito");
while ($query_lista1 = sqlsrv_fetch_array($query_lista)) {
$nome=$query_lista1["nome_inquerito"];
$ativo=$query_lista1["ativo"];
$id_inquerito=$query_lista1["id_inquerito"];
echo "<select>";
echo "<option>$nome</option>";
echo "</select>";
So this code above is functional, but present multiple selects, it's possible that show just one select and inside present all the infos from nome_inquerito on database. 因此,上面的这段代码是有效的,但是可以显示多个选择,有可能只显示一个选择,并且可以在数据库中显示nome_inquerito中的所有信息。
Thanks. 谢谢。
You need to shift <select>
tag before loop. 您需要在循环前移动<select>
标记。
$query_lista = sqlsrv_query($conn, "SELECT * FROM inquerito");
echo "<select>";
while ($query_lista1 = sqlsrv_fetch_array($query_lista)) {
$nome=$query_lista1["nome_inquerito"];
$ativo=$query_lista1["ativo"];
$id_inquerito=$query_lista1["id_inquerito"];
echo "<option>$nome</option>";
}
echo "</select>";
Something like this. 这样的事情。
$query_lista = sqlsrv_query($conn, "SELECT * FROM inquerito");
echo "<select>";
while ($query_lista1 = sqlsrv_fetch_array($query_lista)) {
$nome=$query_lista1["nome_inquerito"];
$ativo=$query_lista1["ativo"];
$id_inquerito=$query_lista1["id_inquerito"];
echo "<option>$nome</option>";
}
echo "</select>";
Try this 尝试这个
write select
outside the while loop
在while loop
外写select
<?php
$query_lista = sqlsrv_query($conn, "SELECT * FROM inquerito");
echo "<select>";
echo "<option value=''>Select Option</option>";
while ($query_lista1 = sqlsrv_fetch_array($query_lista)) {
$nome=$query_lista1["nome_inquerito"];
$ativo=$query_lista1["ativo"];
$id_inquerito=$query_lista1["id_inquerito"];
echo "<option value=".$id_inquerito.">".$nome."</option>";
}
echo "</select>";
?>
$conn = sqlsrv_connect( $serverName, $connectionInfo); $ conn = sqlsrv_connect($ serverName,$ connectionInfo);
/// $SQLquery = NOTE write sql query here
$stmt = sqlsrv_query( $conn, $SQLquery );
if( $stmt === false) {
die( print_r( sqlsrv_errors(), true) );
} else {
Display("SQLquery executed");
}
$result = array();
$fetchLimit = 0; // control the infinite loop
while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {
//echo $row['COLUMN_NAME']. "<br>";
// Loop through each result set and add to result array
$result[] = $row['COLUMN_NAME'];
// results store here = you need to change 'COLUMN_NAME' based on yoru sql query received data name
$fetchLimit++;
if($fetchLimit>60000)
break;
}
echo dropdown( "test", $result, 1000 );
add this function for storing in drop down
function dropdown( $name, array $options, $selected=null )
{
/*** begin the select ***/
$dropdown = '<select name="'.$name.'" id="'.$name.'">'."\n";
$selected = $selected;
/*** loop over the options ***/
foreach( $options as $key=>$option )
{
/*** assign a selected value ***/
$select = $selected==$key ? ' selected' : null;
/*** add each option to the dropdown ***/
$dropdown .= '<option value="'.$key.'"'.$select.'>'.$option.'</option>'."\n";
}
/*** close the select ***/
$dropdown .= '</select>'."\n";
/*** and return the completed dropdown ***/
return $dropdown;
}
Something like this:- 像这样的东西:
<?php
$query_lista = sqlsrv_query($conn, "SELECT * FROM inquerito");
echo "<select>";
echo "<option value=''>Select Nome</option>";
while ($query_lista1 = sqlsrv_fetch_array($query_lista)) {
$nome=$query_lista1["nome_inquerito"];
$ativo=$query_lista1["ativo"];
$id_inquerito=$query_lista1["id_inquerito"];
echo "<option value=".$id_inquerito.">".$nome."</option>";
}
echo "</select>";
?>
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