下拉列表加载数据从数据库PHP javascript
drop down list load data from database php javascript
加载的数据没有在下拉列表中显示
请给我一个解决方案
<html>
<?php
$con = mysqli_connect("localhost","root","","pestpack");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_select_db($con,"pestpack");
$query1 = "SELECT teamname FROM team";
if($rs1 = mysqli_query($con,$query1))
?>
<select>
<option>select a team</option>
<?php
while($row1 = mysqli_fetch_array($rs1))
{
echo '<option value="'.$row['teamname'].'">' . $row['teamname'] . '</option>';
}
?>
</select>
</html>
这样填充,
$con = mysqli_connect("localhost","root","","pestpack");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// mysqli_select_db($con,"pestpack"); No need to select it again, already mentioned in connection
$query1 = "SELECT teamname FROM team";
$rs1 = mysql_fetch_assoc($con,$query1);
echo '<select>';
echo '<option>select a team</option>';
foreach($rs1 as $row) {
echo '<option value="'.$row['teamname'].'">' . $row['teamname'] . '</option>';
}
echo '</select>';
用于在另一个表中插入数据。
<form method="POST" action="team_insert.php"> <!-- Also define it using `echo` -->
<!-- This is generated html select tag-->
<select name="team"> <!-- Define name attribute -->
<option value="team1">team1</option>
<option value="team2">team2</option>
<option value="team3">team3</option>
<option value="team4">team4</option>
</select>
<input type="submit" name="insert" value="Insert" />
</form>
现在在team_insert.php
中写入插入逻辑
if((isset($_SERVER['REQUEST_METHOD'])) &&
($_SERVER['REQUEST_METHOD'] == 'POST') && $_POST['team']) {
$selected_team = $_POST['team'];
// $selected_team variable grap your dropdown value.
// Then write a query logic to insert
}
<html>
<?php
$con = mysqli_connect("localhost","root","","pestpack");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_select_db($con,"pestpack");
$query1 = "SELECT teamname FROM team";
$rs1 = mysql_fetch_array($con,$query1);
?>
<select>
<option>select a team</option>
<?php
foreach ($rs1 as $row) {
echo '<option value='"{$row['teamname']}'">{$row['teamname']}</option>';
}
?>
</select>
</html>
希望这能帮助到一些人!
$conn = mysql_connect("","","");
mysql_select_db("databaseName",$conn);
$sc = mysql_real_escape_string($_SESSION['username']);
$result = mysql_query("SELECT Client_table.Name, Client_table.Client_Id
FROM Client_table, user
WHERE user.username = '$sc'
AND Client_table.Branch = user.area
Order by Name ASC");
//$result = mysql_fetch_assoc($conn, $query1);
echo '<select class="element select large" id="element_17" name="element_17">';
echo '<option> Select Company Name</option>';
while($row = mysql_fetch_array($result)){
echo '<option value="'.$row['Name'].'">' . $row['Name']. '</option>';
}
echo '</select>';
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