我在表格中使用了fgetcsv
,除了最后的额外单元格外,它工作正常。
见http://jobbel.nl/csv.php。我该如何解决这个问题?
这是我的剧本:
<table>
<tr>
<th>Datum</th>
<th>Tijd</th>
<th>Temperatuur</th>
</tr>
<?php
$row = 1;
if (($handle = fopen("test.csv", "r")) !== FALSE) {
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
$num = count($data);
echo "<tr><td>";
$row++;
for ($c=0; $c < $num; $c++) {
echo $data[$c] . "</td><td>";
}
echo "</td></tr>";
}
fclose($handle);
}
?>
</table>
答案 0 :(得分:0)
你回应细胞有点奇怪
将其更改为仅在实际需要时回显<td>
:
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
$num = count($data);
echo "<tr>";
$row++;
for ($c=0; $c < $num; $c++) {
echo "<td>" . $data[$c] . "</td>";
}
echo "</tr>";
}
使用implode()有另一种解决方案,它有点清洁:
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
echo "<tr><td>" . implode('</td><td>', $data) . "</td></tr>";
}
答案 1 :(得分:0)
这应该是......
<?php
$file = file('templog.csv');
foreach($file as $line){
list($datum,$tijd,$temparatuur)=explode(",", $line);
echo "<td>$Date";
echo "<td>$Time";
echo "<td>$Temparature<tr>";
}
?>