Javascript 如何将变量从一个PHP脚本拉到另一个PHP脚本

无法将变量从一个PHP拉到另一个脚本

我有三个不同的文件：adminPage.html、reportScript.php和report.php

html从用户获取变量，并使用AJAX post函数将变量发布到reportScript.php

 $sql = "SELECT * FROM orderlist WHERE NOT (dateOrdered < $startDate OR 
    dateOrdered > $endDate)";

report.php应该从reportScript.php中提取这些已发布的变量，并在SQL函数中使用这些变量，但是，我收到一个错误，指出我有一个未定义的索引：startDate和未定义的索引：endDate，我正在php中实例化这些变量

adminPage.html：

<center><h2> Choose the dates below that you need an order list from: </h2>
</br>
  <form>
    <h2>Start:</h2>
    <input type="date" id ="reportStartDate" name = "startDate">
      </br>
    <h2>End:</h2>
    <input type="date" id ="reportEndDate" name = "endDate">
  </form>
</center>

</br></br>
    <button id="runReportButton" onclick = "runReport()"> Run Report </button>

<script>



function runReport()
{
  var jStartDate;
  var jEndDate;

  jStartDate = document.getElementById("reportStartDate").value;
  jEndDate = document.getElementById("reportEndDate").value;

  /*console.log(jStartDate);
  console.log(jEndDate); */

  $.ajax
  ({
    type: "POST",
    url: "phpScripts/reportScript.php",
    data: {startDate: jStartDate, endDate: jEndDate},
    success: function(response)
      {
        console.log("posted");
       window.open("report.php", "_self");
      }
  });

}

</script>

reportScript.php：

    <?php
    require 'connect.php';

    //posts data to db
    $startDate = $_POST["startDate"];
    $endDate = $_POST["endDate"];

    $sql = "SELECT * FROM orderlist WHERE NOT (dateOrdered < startDate OR 
    dateOrdered > endDate)";

    $result = $conn->query($sql);

    if($result){
    echo "true";
    }

    else{
    echo "false";
    }
    ?>

report.php：

<?php
require 'phpScripts/connect.php';

require 'phpScripts/reportScript.php';

//posts data to db

/*$startDate = $_POST['startDate'];
$endDate = $_POST['endDate'];*/

/*$startDate = '2018-01-01';
$endDate = '2018-08-08'; */

$sql = "SELECT * FROM orderlist WHERE NOT (dateOrdered < '$startDate' OR dateOrdered > '$endDate');";

$result = $conn->query($sql);
//above is reportScript.php, below is pulling list method from order.php
//below works, just needs variables from the reportScript
echo "<ul>";
if($result->num_rows >0)
{
  $i = 0;
  while($row = $result->fetch_assoc()) // this loads database into list, also 
creates array of pricing which someone can pull from later to get total
  {
    echo  "<li style='font-size:15px'>".$row["drinkName"]. ", Date Ordered: " 
.$row["dateOrdered"] . ",Cost: " .$row["drinkCost"] . "</li>";
    echo "</br>";

  $i = $i+1;
  }
}else {
  echo "<p> you're a dummy and you did this wrong </p>";
}
echo "</ol>";


?>

您在reportScript.php中的变量中忘记了美元符号$

 $sql = "SELECT * FROM orderlist WHERE NOT (dateOrdered < $startDate OR 
    dateOrdered > $endDate)";

---

此语句也容易受到攻击。

根据@Ralf的一些建议，我将reportScript.php和report.php合并在一起，并在打开时使用$\u GET语句将日期变量放入URL中。这样，查询就不会被放置两次，变量也会被保存。

谢谢您的帮助！美元符号肯定是我错过的东西，但是它并没有解决问题，我仍然会出现同样的错误。我认为问题在于你的ajax中的window.open。您正在启动另一个线程，当您需要reportScript.php脚本时，$\u POST数组不再分配。也许您应该在ajax请求和php工作输出中完成所有必须完成的工作。非常感谢您的帮助。我结合了report.php和reportScripts.php，本质上是将AJAX函数直接发布到report.php。但是，我不确定如何在不启动报表的情况下打开report.php窗口，因此会出现相同的错误。