Question

我想在我的数据库中输入用户测量值，但我需要先将其格式化，但结果字符串似乎是空的这是我到目前为止所拥有的..

$data = $_POST['mesurement'];
$users_mesurement;
//if data has . 
  if (strpos($data,'.') !== false) {
   //explode data
   $testA = explode(".",$data);
    // if testA[0] str is more than 2
    if (strlen($testA[0]) > 2){
    // if testA[1] str is below 1
     if (strlen($testA[1]) < 2){
      $users_mesurement = $data . "0";
     // if testA[1] str is above 1
     }else{
      $users_mesurement = $data;
     }
    // has . but below 2
    }else{
     // if testA[1] str is below 1
     if (strlen($testA[1]) < 2){
      $users_mesurement = " " . $data . "0";
     // if testA[1] str is above 1
     }else{
      $users_mesurement = " " . $data;
     }
//if does not have . 
  }else{
    //if is more than 2
    if (strlen($data) > 2){
     $users_mesurement = $data . ".00";

    //if less than 2
    }else{
     $users_mesurement = " " . $data . ".00";

    }
  }

可能的$数据类型是
NNN.NN
NNN.N
NNN
NN.NN
NN.N
NN
N =任意数字
$ users_mesurement结果必须是＆＃34; NNN.NN＆＃34;或者＆＃34; NN.NN＆＃34;

好的，我现在有了这个

//if data has . 
  if (strpos($data,'.') !== false) {
   //explode data
   $testA = explode(".",$data);
   $part1 = sprintf("%03d", $testA[0]);
   $part2 = sprintf("%02d", $testA[1]);
   $users_mesurement = $part1 . "." . $part2 . " cm"; 
//if does not have . 
  }else{
   $part1 = sprintf("%03d", $data);
   $users_mesurement = $part1 . "." . "00" . " cm"; 
  }

并且当我输入99.5时它除了一部分它起作用它输出099.05我需要它是099.50

   $part2 = str_pad($testA[1], 2 , "0");

现在这一切都很好了

mysql数据输入格式不起作用

0 个答案: