Question

我试图将一个PHP变量（在本例中为$ beg）输入到mySQL查询中，但它返回一个空数组结果。数据库中字段的类型是整数。当我输入实际值而不是变量时，我得到正确的结果。怎么了？

    $beg = time()-5000;
    settype($beg, "integer");

    $result = mysql_query('SELECT * FROM records WHERE time>=$beg ORDER BY time ASC');

    $statusdata = array();
    while ($row = mysql_fetch_array($result)) {
    array_push($statusdata, $row["status"]);
    }

Answer 1

确保在字符串中使用$ variables时使用双引号。

    $result = mysql_query("SELECT * FROM records WHERE time>= $beg ORDER BY time ASC");

Answer 2

您应该使用预准备语句而不是mysql_query。

$beg = time()-5000;
settype($beg, "integer");

$db = new mysqli("host","user","pw","database");
$stmt = $db->prepare("SELECT status FROM records WHERE time>=? ORDER BY time ASC");
$stmt->bind_param('i', $beg);
$stmt->execute();

$stmt->store_result();
$stmt->bind_result($status);

$statusdata = array();
while($stmt->fetch())
{
  array_push($statusdata, $status);
}

$stmt->close();

Answer 3

更改行

$result = mysql_query("SELECT * FROM records WHERE time>=$beg ORDER BY time ASC");

您必须使用双引号字符串来放置变量。

Answer 4

更改您的查询

$result = mysql_query(" SELECT * FROM records WHERE time >= $beg ORDER BY time ASC ");

您不能在单引号内使用变量。

Answer 5

尝试这种方法，我经常使用：

$beg = time() - 5000;
$query = sprintf("SELECT * FROM %s WHERE time >= '%o' ORDER BY %s ASC", "records", $beg, "time");
$result = mysql_query($query);

记住，time（）结果是整数，你不需要将他设置为整数

mySQL查询中的PHP整数变量

5 个答案: