解决php fgetcsv 读取csv文件数据不完整问题

作者:袖梨 2022-06-24

在windows其它版本中

 代码如下 复制代码

# Open the File.
if (($handle = fopen("test.csv", "r")) !== FALSE) {
    # Set the parent multidimensional array key to 0.
    $nn = 0;
    while (($data = fgetcsv($handle, 0, ",")) !== FALSE) {
       
//print_r($data);
        # Count the total keys in the row.
        $c = count($data);
        # Populate the multidimensional array.
        for ($x=0;$x<$c;$x++)
        {
            $csvarray[$nn][$x] = $data[$x];
        }
        $nn++;
    }
    # Close the File.
    fclose($handle);
}
//print_r($csvarray);

这个代码没有任何问题,然后我放到了linux中发现有为空的字段了。

问题解析出来的数据不完整,有为空的字段
网上查了下说是在php5.2.8 中存在bug
解决办法是使用自定义函数

 代码如下 复制代码

function __fgetcsv(& $handle, $length = null, $d = ',', $e = '"') {
     $d = preg_quote($d);
     $e = preg_quote($e);
     $_line = "";
     $eof=false;
     while ($eof != true) {
         $_line .= (empty ($length) ? fgets($handle) : fgets($handle, $length));
         $itemcnt = preg_match_all('/' . $e . '/', $_line, $dummy);
         if ($itemcnt % 2 == 0)
             $eof = true;
     }
     $_csv_line = preg_replace('/(?: |[ ])?$/', $d, trim($_line));
     $_csv_pattern = '/(' . $e . '[^' . $e . ']*(?:' . $e . $e . '[^' . $e . ']*)*' . $e . '|[^' . $d . ']*)' . $d . '/';
     preg_match_all($_csv_pattern, $_csv_line, $_csv_matches);
     $_csv_data = $_csv_matches[1];
     for ($_csv_i = 0; $_csv_i < count($_csv_data); $_csv_i++) {
         $_csv_data[$_csv_i] = preg_replace('/^' . $e . '(.*)' . $e . '$/s', '$1' , $_csv_data[$_csv_i]);
         $_csv_data[$_csv_i] = str_replace($e . $e, $e, $_csv_data[$_csv_i]);
     }
     return empty ($_line) ? false : $_csv_data;
}

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