正在尝试根据返回值覆盖现有数据,只想获取'location‘以更新来自CSV导入的新信息。有一个带有Oldlocation的复制表(为了测试,全部设置为DW,因此可以看到任何更改),看起来应该很简单...:
$file = "allinv.CSV";
if(($handle = fopen($file, "r")) !== FALSE)
{ fgetcsv($handle);
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){
$num = count($data);
for ($c=0; $c < $num; $c++) {
$col[$c] = $data[$c];}
$col1 = $col[1];
$sql = "SELECT Oldlocation FROM invbkup WHERE VIN = '$col1'"; ///works fine, gets old location for each record
$result = mysqli_query($con, $sql);
$row = mysqli_fetch_array($result);
$whl= $row['Oldlocation'];
$col8 = $col[73];
$query = "UPDATE allinv SET location = '$col8' WHERE $whl = 'DW' ";
echo $col1, $col8, $whl, "</br>";////can see all the info but get no changes in database.
$a= mysqli_query($con, $query);
}
fclose($handle);
}
发布于 2016-09-19 21:42:19
不要紧,这是一个简单的问题,我把它放在一个if语句中:
if($whl = 'DW'){
$query = "UPDATE allinv SET location = '$col8' WHERE VIN ='$col1'";
$a= mysqli_query($con, $query);}
有时你自己把问题复杂化了,休息一下,喝点咖啡,试着不去想那么多。
https://stackoverflow.com/questions/39598545
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