此代码有什么问题。如果表为空,我不能进入插入函数,它总是在进行更新。谁能帮我解决这个问题?
while ($row = $result->fetch_object()) {
if ($result->num_rows > 0) {
$updatequery = sprintf("UPDATE stat_mailings SET cat_id='%s', mailing_name='%s', mailing_unique_id='%s', segment_id='%s',
campaign_id='%s', landing_page='%s', total_sent='%s' WHERE mailing_id='%s'",
$row->category_id,
mysqli_real_escape_string($mysqli, $row->mailing_naam),
$row->unique_id,
$row->mailing_segment,
mysqli_real_escape_string($mysqli, $row->utm_campaign),
mysqli_real_escape_string($mysqli, $row->landing_page),
$row->mailing_total_sent,
$row->id
);
$mysqli->query($updatequery);
echo $error = $mysqli->error;
} else {
$insertquery = sprintf("INSERT INTO stat_mailings SET mailing_id='%s', cat_id='%s', mailing_name='%s', mailing_unique_id='%s', segment_id='%s',
campaign_id='%s', landing_page='%s', total_sent='%s'",
$row->id,
$row->category_id,
mysqli_real_escape_string($mysqli, $row->mailing_naam),
$row->unique_id,
$row->mailing_segment,
mysqli_real_escape_string($mysqli, $row->utm_campaign),
mysqli_real_escape_string($mysqli, $row->landing_page),
$row->mailing_total_sent
);
$mysqli->query($insertquery);
echo $error = $mysqli->error;
}
答案 0 :(得分:4)
Mysql具有完全用于此目的的强大功能 - ON DUPLICATE查询 它看起来像
INSERT INTO t SET f1=val,f2=val2 ON DUPLICATE KEY UPDATE f1=val,f2=val2
等等。
只确保您在关键字段上有一个独特的索引
因此,您可以摆脱SELECT查询并进行结果检查。