Question

I want to export CSV file to JSON file,require is read value at recommend field and write to JSON file. Exapmle : IN row 1,recomnend =2 will read the row that id=2 and print to JSON file that values : id,product_id,title.

$json = array();

        $json['id'] = $row['id'];
        $json['product_id'] = $row['product_id'];
        $json['title'] = $row['title'];
        $json['outline'] = $row['outline'];
        $recom[] = $json;  
print json_encode($recom);

http://ns0.upanh.com/b6.s33.d3/3a3ebb93a87c5703f673b399d5613ec2_50639320.untitled.png

Answer 1

I don’t know of a different way of doing it, but the following will work. Using this method means that you have to use the column index, not the name of it. This shouldn’t be a problem if you know the column index of the values you need, but I just thought I would mention:

$fp = fopen('test.csv','r') or die("**! can't open file\n\n");
while($csv_line = fgetcsv($fp,1024)) {
    $id = $csv_line[0];
    $product_id = $csv_line[1];
    $title = $csv_line[2];
    $outline = $csv_line[3];        
}
fclose($fp) or die("**! can't close file\n\n");

Now, because you will have multiple rows, I would recommend the following to save them into one JSON object:

$fp = fopen('test.csv','r') or die("**! can't open file\n\n");
$i = 0;
while($csv_line = fgetcsv($fp,1024)) {
    $i++;
    $json['json_'.$i]['id'] = $csv_line[0];
    $json['json_'.$i]['product_id'] = $csv_line[1];
    $json['json_'.$i]['title'] = $csv_line[2];
    $json['json_'.$i]['outline'] = $csv_line[3];        
}
$json['total_lines'] = $i;
print json_encode($json);
fclose($fp) or die("**! can't close file\n\n");

With this method, you can save each row as a sub-JSON Object, making it possible to pull the total number and parse through the objects to pull out the rows.