Question

我无法弄清楚为什么这段代码不起作用。我试图创建一个临时表并将一些数据拉入其中，然后过滤掉一些数据。但我似乎无法回应数据它只是回应列名称。这是代码：

if (!$conn) {
    echo "<a href='getschedule.php'><button>Go Back</button></a>";
    die("Connection failed: " . mysqli_connect_error());

}else{
    echo "Connected successfully<br><br>";
}
$seasontemp = "CREATE TEMPORARY TABLE seasontemp (
id int NOT NULL,
firstname varchar(255),
lastname varchar(255),
number varchar(255),
address varchar(255),
plan_start date NOT NULL,
plan_comp int(11),
plan_skip int(11),
trim_start date,
trim_comp int(11),
trim_skip int(11),
spray_start date,
spray_comp int(11),
PRIMARY KEY(id)
)";

mysqli_query($conn, $seasontemp) or die ("Sql error : ".mysqli_error($conn));

$insertseason = "INSERT INTO seasontemp
    (id, plan_start, plan_comp, plan_skip, trim_start, trim_comp, spray_start, spray_comp)
SELECT id, plan_start, plan_comp, plan_skip, trim_start, trim_comp, spray_start, spray_comp
FROM services WHERE plan='17'";

mysqli_query($conn, $insertseason) or die ("Sql error : ".mysqli_error($conn));

$seasonids = "SELECT 'id', 'plan_start' FROM seasontemp";
$r1 = mysqli_query($conn, $seasonids ) or die ("Sql error : ".mysqli_error($conn));
if(mysqli_num_rows($r1) > 0){
$start = mysqli_fetch_assoc($r1);
$start_date = $start['plan_start'];
echo $start_date;

}

mysqli_close($conn);

Answer 1

"SELECT 'id', 'plan_start' FROM seasontemp";

将其替换为：

"SELECT `id`, `plan_start` FROM seasontemp"; // wrong quotes added here.

Answer 2

将"SELECT 'id', 'plan_start' FROM seasontemp";替换为"SELECT id, plan_start FROM seasontemp";

注意：删除单引号的字段，否则使用＆＃34;`＆＃34;不是＆＃34;＆＃39;＆＃34;。

Answer 3

您检查您的插入查询是否正确？

$insertseason = "INSERT INTO seasontemp
(id, plan_start, plan_comp, plan_skip, trim_start, trim_comp, spray_start, spray_comp)
SELECT id, plan_start, plan_comp, plan_skip, trim_start, trim_comp, spray_start, spray_comp
FROM services WHERE plan='17'";

我认为错误的插入查询。纠正它

无法使用mysqli从数据库获取数据

3 个答案: