PHP仅在csv的一列中替换数据 [英] PHP replace data only in one column of csv

问题描述

我只需要替换csv中第三列中的值即可.而且没有别的地方. 我的csv的第3列仅包含"J"或"N"，我需要将其更改为"5"或"0".但是现在我改变了，在整个文件中总是找到所有N和J#.但是我只需要在第三栏中进行更新.

I need to replace the values only from the third column in my csv. And no where else. Column 3 of my csv contains only a 'J' or a'N' and i need to change them to '5' or '0'. But right now i it changes always all found N's and J#s in the whole file. But i need it only updated in the third column.

我使用以下代码打开并重写我的csv.但是它总是将找到的所有字母写为5或0.

I use the following code to open and rewrite my csv. But it writes always all found Letters to 5 or 0.

$csvfile = 'csvfile.csv';
if (file_exists($csvfile)) {
        unlink($csvfile);
    echo "Alte $csvfile entfernt!";
} else {
    echo "Keine alte $csvfile vorhanden";
}

$row = 0;
$fp = fopen('csvfile.csv', 'w');
if (($handle = fopen("oldCSV.csv", "r")) !== FALSE) {
    while (($data = fgetcsv($handle, 200000, ",")) !== FALSE) {
    $num = count($data);
    $dataold = ['J', 'N'];                                                            
    $datanew = ['5', '0'];
    echo "<p> $num Felder in Zeile $row: <br /></p>\n";
    $row++;
    for ($c=0; $c < $num; $c++) {
        fputs($fp, str_replace($dataold, $datanew, $data[$c] . "\n"));
    }
  }
}

fclose($fp);

推荐答案

while ($data = fgetcsv($handle)) {

    $data[2] = str_replace(['J', 'N'], ['5', '0'], $data[2]);

    fputcsv($fp, $data);

    echo "<p> ".count($data)." Felder in Zeile ".$row++.": <br /></p>\n";
}

编辑

1. 我从while中删除了!== FALSE部分，每次都不将结果与false进行比较，最终fgetcsv()本身将返回false并停止循环
2. 折叠了str_replace及其参数数组(仅出于可读性)
3. 删除了for循环以仅更改第3个([2])列
4. 摆脱$num变量

1. I removed !== FALSE part from while to not compare result with false each time, finally fgetcsv() will return false itself and stop the loop
2. collapsed str_replace and its' arguments-arrays (for readability only)
3. removed for loop to change only 3rd ([2]) coloumn
4. got rid of $num variable

这篇关于PHP仅在csv的一列中替换数据的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！