我的表设置如下图所示。
当我尝试运行以下代码将值插入数据库时,我收到错误:
失败:插入
Betfairodds
(Horse
,Back
,Lay
,TimeformTR
)VALUES('Intrepid','5.5','5.9', '0')
有人能帮忙,因为我试图调试代码。
//loop through each individual card
foreach ($getdropdown2 as $dropresults) {
$horse = preg_replace('/\h*[^ a-zA-Z].*$/m', '', trim($dropresults->childNodes->item(8)->textContent));
$back = trim(GetBetween($dropresults->childNodes->item(18)->textContent, 'Back', '£'));
$lay = trim(GetBetween($dropresults->childNodes->item(20)->textContent, 'Lay', '£'));
$sql = "INSERT INTO `Betfairodds` (`Horse`,`Back`,`Lay`,`TimeformTR`)VALUES( '$horse','$back', '$lay', '0')";
$res = mysqli_query($db, $sql);
if (!$res) {
echo PHP_EOL . "FAIL: $sql";
trigger_error(mysqli_error($db), E_USER_ERROR);
}
}
答案 0 :(得分:0)
我从0中删除了引号'
,因为它在架构中定义为int
,当然在VALUES
之前添加了空格..
$sql = "INSERT INTO `Betfairodds` (`Horse`,`Back`,`Lay`,`TimeformTR`) VALUES( '$horse','$back', '$lay', 0)";
答案 1 :(得分:-1)
你的陈述是错误的。您不应该在数据字段上放置单引号。所以应该是这样的:
$sql = "INSERT INTO `Betfairodds` (Horse,Back,Lay,TimeformTR)VALUES( '$horse','$back', '$lay', '0')";