这是我的表:
Place_Id | Value_Name| Value
-------------------------------
001 | Name1 | Value1
-------------------------------
001 | Name2 | Value2
-------------------------------
001 | Name3 | Value3
-------------------------------
002 | Name1 | Value4
-------------------------------
002 | Name2 | Value5
如何回显Place_Id
001的所有Value_Name
的列表?
我试过这个:
<?php
$query2 = ("SELECT * FROM table WHERE Place_Id = 001");
if ($statement2 = $db_conn_pdo->prepare($query2))
{
$statement2->execute();
while ($row2 = $statement2->fetch(PDO::FETCH_ASSOC))
{
$output2 = $row2['Value_Name'];
}
}
echo $output2;
?>
并且它只返回最后一个"Value_Name"。
这样的东西应该能在中工作
$query = $con ->query
("
SELECT Value_Name FROM [table] WHERE Place_Id = '001'
");
$valueName = array();
while($row = $query->fetch_object())
{
$valueName[] = $row;
}
foreach($valueName as $Value)
{
echo $Value->Value_Name;
}
这是最终版本,感谢@Alexander Ravikovich和@McNoodles:
$query = $con ->query
("
SELECT Value_Name FROM [table] WHERE Place_Id = '001'
");
while($row = $query->fetch_object())
{
echo $row['Value_Name'];
}
// the second loop is not needed as @Alexander Ravikovich suggested
假设表名是抖动(因为您没有提到它)
SQL查询:SELECT Value_Name FROM jiggles WHERE jiggles.place_id = [the id you want to know]
通过PHP执行此操作,并在的某个位置回显结果
(Select中的*也可以,只要只回显value_name属性即可)