Question

我想将csv文件上传到sql表中。这是我的代码：

<?php

include "connection.php"; //Connect to Database

$deleterecords = "TRUNCATE TABLE axioma_ordini"; //empty the table of its current records
mysql_query($deleterecords);

//Upload File
if (isset($_POST['submit'])) {
    if (is_uploaded_file($_FILES['filename']['tmp_name'])) {
        echo "<h1>" . "File ". $_FILES['filename']['name'] ." uploaded successfully." . "</h1>";
        echo "<h2>Displaying contents:</h2>";
        readfile($_FILES['filename']['tmp_name']);
    }

    //Import uploaded file to Database
    $handle = fopen($_FILES['filename']['tmp_name'], "r");

    while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
        $import="INSERT into axioma_ordini(`id`, `Famiglia`, `Cod Cliente`, `Ragione Sociale`, Articolo, Descrizione, Tipo, `Valore Euro`, Valuta, `Valore in Valuta`, Cambio, `Mese Competenza`, Anno, `Cir Medio`, Giorni, Satellite, `Satellite per Risorsa`, NumeroDocumento, `Data Doc`, `Nome Nave`, `Modem S/N`, `Burst Fwd`, `Burst Rtn`, Condivisione, `Cir Fwd`, `Cir Rtn`, `Cir Tot`, Business, Tecnologia,`Data Scadenza Ordine`, Network, Transponder, Nazione) values('','$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]','$data[7]','$data[8]','$data[9]','$data[10]','$data[11]','$data[12]','$data[13]','$data[14]','$data[15]','$data[16]','$data[17]','$data[18]','$data[19]','$data[20]','$data[21]','$data[22]','$data[23]','$data[24]','$data[25]','$data[26]','$data[27]','$data[28]','$data[29]','$data[30]','$data[31]')";

        mysql_query($import) or die(mysql_error());
    }

    fclose($handle);

    print "Import done";

    //view upload form
}else {

    print "Upload new csv by browsing to file and clicking on Upload<br />\n";

    print "<form enctype='multipart/form-data' action='upload.php' method='post'>";

    print "File name to import:<br />\n";

    print "<input size='50' type='file' name='filename'><br />\n";

    print "<input type='submit' name='submit' value='Upload'></form>";

}

?>

它导入成功，但是在第一列：Famiglia（和id工作）。其余的没有值或0值。我要导入的文件包含数据。此外，在famiglia中，它是varchar 5，它从下一列获取第一个数字。因此，它不是商业，而是商业; 8（8是下一栏的第一个数字）。

有谁知道问题出在哪里？

谢谢！

Answer 1

我发现了问题所在。这是一个错误的分隔符：而不是;

（$ data = fgetcsv（$ handle，1000，“，”）

应该是

（$ data = fgetcsv（$ handle，1000，“;”）

也许这有助于某人...

Sql表只导入第一列

1 个答案: