Question

我想将所有数据从给定值1输出到value2。我的代码正在运行，但它只显示一个数据。

<?php 

  mysql_connect("localhost","root","") or die("NO connection");
  mysql_select_db("ci")or die("No DATABASE");

  $value1 =$_POST['value1'];
  $value2 =$_POST['value2'];
  $output="";

  if (isset($value1,$value2)){

    $result=mysql_query("SELECT * FROM slogs WHERE stime BETWEEN 
            '". $value1 ."'AND '". $value2 ."' ");
    $count=mysql_num_rows($result);

    if($count==0){
      $output='NO DATA IN DATABASE';
    }

    if ($count>1){
      while($row=mysql_fetch_array($result))
      { 
        $TIME=$row['stime'];
        $DT=$row['DT'];
        $FN=$row['FN'];
        $LN=$row['LN'];
      }
    }

  }

?>

Answer 1

首先像这样更改您的查询

mysql_query("SELECT * FROM `slogs` WHERE stime BETWEEN 
 '". $value1 ."' AND '". $value2 ."'") or die(mysql_error());

然后尝试这种方式;在while循环内，每当您将数据overriding添加到相同的变量时

while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
 {
    printf("Time: %s  Date: %s FN: %s LN: %s", $row["stime"],$row["DT"],$row["FN"],$row["LN"]); 
}`

如何从一个表输出所有数据

1 个答案: