我将如何执行两个查询与两个不同的按钮


how i will execute two query with two different button

这是html文件

<label>Search:</label>
<input id="dontprint" type="text" style="" name="ssrch1">   
<input type="submit" Value="Search" id="dontprint"  class ="btn btn-success" name="btnssrch1">  
<label>Date From:</label>
<input type="text"id="jQueryDatePicker1" name="date1">
<label> To: </label>
<input type="text" id="jQueryDatePicker2" name="date2">
<input type="submit" Value="Search" id="dontprint" class="btn btn-success" name="btnssrch2">

在这个函数中,如果你点击btnssrch,它将运行查询,但这不是我点击btnssrch后查询没有运行的情况。第二个查询正在运行,但这不是我应该做什么?

if(isset($_POST["btnssrch1"])) {        
    $cno1 = mysqli_real_escape_string($link, $_POST["ssrch1"]);   
    $p = mysqli_query($link,"select * from patient where CONCAT(PatientId,' ', FirstName,' ',MiddleName,' ',LastName) like '%$cno1%'");
}

在这个函数中,查询正在执行。

 if(isset($_POST["btnssrch2"])) {
     $date1=mysqli_real_escape_string($link,$_POST["date1"]);
     $date2=mysqli_real_escape_string($link,$_POST["date2"]);
     $p = mysqli_query($link,"select * from patient where Date BETWEEN '" . $date1 . "' AND  '" . $date2 ."'");
 } else {
     $p=mysqli_query($link,"select * from patient");
 }
 while($r=mysqli_fetch_array($p)){

你应该使用<button>标签而不是<input type=submit>,当你点击一个特定的按钮时,这个按钮将被设置,现在你可以使用isset函数来检查。按钮的其余部分将不会设置,直到您单击它。如:

<!doctype html>
<html>
<head> 
</head>
<body>
<form action="<?php htmlspecialchars($_SERVER['PHP_SELF']) ; ?>" method="post">
<button type="submit" name="button1"> firstButton </button>
<button type="submit" name="button2" >secondButton </button>
</form>
<?php
if ($_SERVER['REQUEST_METHOD'] == "POST") {
    if (isset($_POST["button1"])) {
        echo "button 1 is set.<br>"; 
        var_dump(isset($_POST["button2"])); // var_dump will return the data type and the value.
    } elseif (isset($_POST["button2"])) {
        echo "button 2 is set.";
    } else {
        echo "no button is set.";
    }
}
?>
</body>
</html>

当您单击第一个按钮时,表示button1已设置。当你点击button2时,它说按钮2被点击了

或者你可以使用两个<form>元素和两个隐藏输入,就像其他人建议的那样。