无法生成.csv文件:PHP

Question

实际上我不知道如何生成.csv文件.我已经完成了各种电子笔记,但仍然没有得到.

我想做的事？

我写了一段代码来回显 " partner_name "和" link ",但这些代码需要放入formDetails.csv文件中.

$result = mysql_query(
    "SELECT form_id, partner_name 
     FROM `ef_kabaadkhana`.`ef_form_master_v1` 
     WHERE active = 1 AND listed = 1;"
);

while ($formDetails = mysql_fetch_assoc($result)) {
    $partner_name       = $formDetails['partner_name'];
    $formId             = $formDetails['form_id'];
    $slugifyPartnername = slugify($partner_name);
    $partner_link       = "http://myforms.com/form/{$formId}/{$slugifyPartnername}";
    echo $partner_name . '=>' . $partner_link, "\n";
}

function slugify($text) {
    $text = preg_replace('~[^\\pL\d]+~u', '-', $text);
    $text = trim($text, '-');
    if (function_exists('iconv'))
        $text = iconv('utf-8', 'us-ascii//TRANSLIT', $text);
    $text = strtolower($text);
    $text = preg_replace('~[^-\w]+~', '', $text);
    if (empty($text))
        return 'n-a';
    return $text;
}

它能做什么？

它只需要从ef_form_master_v1表中获取form_id和partner_name .的$partnerName和$partnerlink被定义,并相应地显示.

而是呼应$partnerName和$partnerlink,我希望把这些阵列的参数在名为CSV文件formDetails.csv.

Answer 1

该文档上fputcsv是很清楚,但我会在下面提供了一个例子:

$handle = fopen('/path/to/formDetails.csv', 'w');

while(<your query loop>) {
    $write = array($partner_name, $partner_link);
    fputcsv($handle, $write);
}

fclose($handle);