我正在使用LOAD DATA INFILE替換一個短腳本,以便使用PHP將CSV快速插入到MySQL數據庫中。在構建MySql中使用PHP變量變量INSERT
我完全難住,並希望得到一些幫助 - 我一直在看這個小時,只是困惑自己!
問題是我不知道CSV中的列數,也不想定義它們,因爲它們可能會發生變化。
從CSV獲得標題:
// Get the first column of the file to create the column headers
$fp = fopen($upload_folder . $tempName . '.csv', 'r');
$frow = fgetcsv($fp, 100000, ',');
// Set import strings
$columns = '';
$data_string = '';
$col_count = 0;
foreach($frow as $column) {
// Build import strings
$columns .= "`$column` varchar(50)";
$data_string .= "'$" . "data[" . $col_count . "]', ";
// Now increment the column count
$col_count++;
}
進口:(我已經回顯的這個,因爲我不希望它運行!)
while (($data = fgetcsv($fp, 100000, ",")) !== FALSE) {
echo "INSERT INTO nav_data_" . $tempName . " (" . $columns . ") VALUES (" . $data_string . ") ";
}
問題: 雖然列名在WHILE語句中調用的第二個$ data [0],$ data [1]沒有被解析,所以我得到:
VALUES ('$data[0]', '$data[1]', '$data[2]', '$data[3]', '$data[4]', '$data[5]', '$data[6]' ... '$data[25]')
在哪裏,因爲我想從$數據= fgetcsv($ fp的,100000,「」)的輸出...
VALUES ('1', 'Farming', 'Tree', 'Tractor', '345', 'etc.')
我知道這是用PHP變量變量,但對我來說我的生活中做不能解決這個問題。
例CSV數據導入:(根據要求)
Document Type No_ Order Date Shipment Date Sell-to Customer No_ Sell-to Customer Name Bill-to Customer No_ Bill-to Name Ship-to Code Order Status Customer Disc_ Group Salesperson Code Entered by Taken at External Document No_ Taken By Sample Payment Method Code Currency Code Payment Terms Code Amount O/S Amount Ex VAT (LCY) No_ Order Lines No_ Complete Order Lines Delivery Date
Order SO006906 02/12/14 15/04/15 C001293 Amazon C001293 Amazon DEL Insufficient Credit HS SARAH EMAIL 0 ACCOUNT 30D 2313 2313 15 11
Order SO007198 13/01/15 01/04/15 C000871 Amazon C000871 Amazon DEL Awaiting Order Ack. SUSP SARAH FF15 RESERVE SUSP 0 ACCOUNT 30DS2.5/30 688 688 4 0
Order SO008292 01/05/15 19/10/15 C003075 Amazon C003075 Amazon DEL Proforma Ready To Ship ZP SARAH REP 0 PROFORMA PROFORMA 484.8 484.8 21 0
你可以從csv文件發佈一些你的數據,幾行嗎? – mandza
我已經將一些數據添加到原始帖子mandza - 它太長的評論 - 感謝您看這個! – Giles