因此,目前我正在尝试制作一个数据库导出脚本,我得到的是:
<?php
$con = mysqli_connect("SERVER","USER","PASSWORD","DATABASE");
$query = mysqli_query($con,"SELECT * FROM `TABLE`;");
unlink('testfile.csv');
$fp = fopen("testfile.csv","w");
fputcsv($fp,array("Col1","Col2","Col3"),";");
while ($list = mysqli_fetch_array($query)){
fputcsv($fp,$list,";");
}
fclose($fp);
?>这似乎返回了"Col1 Col1 Col2 Col2 Col3 Col3"
而fputcsv($fp,array($list[0],$list[1],$list[2]),";");似乎工作得很好。是否有一种方法可以避免这种解决方案,并由原始的$list数组来完成?
发布于 2014-07-11 07:26:05
尝试:
$list = mysqli_fetch_array($query, MYSQLI_NUM);这应该以数字数组的形式返回数组,而不是默认返回的混合映射。在输出上也尝试print_r,以确保它是您所期望的。
http://uk3.php.net/manual/en/mysqli-result.fetch-array.php
https://stackoverflow.com/questions/24692069
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