我想用php上传一个csv文件。上传文件后,我想显示CSV文件的数据。我想举一个如何完成这项任务的例子。
尽管您可以轻松找到如何使用php处理文件上传的教程,并且有一些功能(手动)可以处理CSV,但是我将发布一些代码,因为几天前我从事一个项目,其中包括一些代码,您可以采用…
HTML:
<table width="600"> <form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post" enctype="multipart/form-data"> <tr> <td width="20%">Select file</td> <td width="80%"><input type="file" name="file" id="file" /></td> </tr> <tr> <td>Submit</td> <td><input type="submit" name="submit" /></td> </tr> </form> </table>
PHP:
if ( isset($_POST["submit"]) ) { if ( isset($_FILES["file"])) { //if there was an error uploading the file if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { //Print file details echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />"; //if file already exists if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { //Store file in directory "upload" with the name of "uploaded_file.txt" $storagename = "uploaded_file.txt"; move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $storagename); echo "Stored in: " . "upload/" . $_FILES["file"]["name"] . "<br />"; } } } else { echo "No file selected <br />"; } }
我知道必须有一种更简单的方法来执行此操作,但是我读取了CSV文件并将每个记录的单个单元格存储在二维数组中。
if ( isset($storagename) && $file = fopen( "upload/" . $storagename , r ) ) { echo "File opened.<br />"; $firstline = fgets ($file, 4096 ); //Gets the number of fields, in CSV-files the names of the fields are mostly given in the first line $num = strlen($firstline) - strlen(str_replace(";", "", $firstline)); //save the different fields of the firstline in an array called fields $fields = array(); $fields = explode( ";", $firstline, ($num+1) ); $line = array(); $i = 0; //CSV: one line is one record and the cells/fields are seperated by ";" //so $dsatz is an two dimensional array saving the records like this: $dsatz[number of record][number of cell] while ( $line[$i] = fgets ($file, 4096) ) { $dsatz[$i] = array(); $dsatz[$i] = explode( ";", $line[$i], ($num+1) ); $i++; } echo "<table>"; echo "<tr>"; for ( $k = 0; $k != ($num+1); $k++ ) { echo "<td>" . $fields[$k] . "</td>"; } echo "</tr>"; foreach ($dsatz as $key => $number) { //new table row for every record echo "<tr>"; foreach ($number as $k => $content) { //new table cell for every field of the record echo "<td>" . $content . "</td>"; } } echo "</table>"; }
因此,我希望这会有所帮助,这只是一小段代码,并且我尚未对其进行测试,因为我使用的代码略有不同。评论应解释一切。