我正在将数据导出到.csv
文件并且它运行正常,但我有一个小问题。我从表格中提取name
和gender
但是为了性别,我将id
保存在我的数据库中(即1 = Male
,2 = Female
)。我的下面的代码给了我性别的ID,我该如何解决?男性返回1,女性返回2:
$rows = mysql_query("SELECT `name`, `gender` FROM TABLE");
while ($row = mysql_fetch_assoc($rows)) {
fputcsv($output, $row);
}
答案 0 :(得分:6)
试试这个:
$rows = mysql_query("SELECT `name`, `gender` FROM TABLE");
while ($row = mysql_fetch_assoc($rows)) {
if($row['gender'] == 1) {
$row['gender'] = 'Male';
} else {
$row['gender'] = 'Female';
}
// Or ternary condition
// $row['gender'] = ($row['gender'] == 1 ? 'Male' : 'Female');
fputcsv($output, $row);
}
答案 1 :(得分:2)
<?php
$db_record = 'yourRecod';
// optional where query
$where = 'WHERE 1 ORDER BY 1';
// filename for export
$csv_fileName = 'db_export_'.$db_record.'_'.date('Y-m-d').'.csv';
// database variables
$hostname = "localhost";
$user = "yourUserName";
$password = "yourPassword";
$database = "yourDataBase";
// Database connecten voor alle services
mysql_connect($hostname, $user, $password)
or die('Could not connect: ' . mysql_error());
mysql_select_db($database)
or die ('Could not select database ' . mysql_error());
$csv_export = '';
$query = mysql_query("SELECT * FROM ".$db_record." ".$where);
$field = mysql_num_fields($query);
// create line with field names
for($i = 0; $i < $field; $i++) {
$csv_export.= mysql_field_name($query,$i).';';
}
$csv_export.= '';
while($row = mysql_fetch_array($query)) {
// create line with field values
for($i = 0; $i < $field; $i++) {
$csv_export.= '"'.$row[mysql_field_name($query,$i)].'";';
}
$csv_export.= '';
}
// Export the data and prompt a csv file for download
header("Content-type: text/x-csv");
header("Content-Disposition: attachment; filename=".$csv_fileName."");
echo($csv_export);
?>
我给你完整的示例代码,我用它来使用PHP将MySql数据导出到.CSV
答案 2 :(得分:2)
使用SQL中的IF statement
来最小化PHP逻辑:
SELECT `name`, IF(gender=1,'Male','Female') as 'gender' FROM TABLE