在Javascript中嵌入PHP

Question

I need to embed PHP into Javascript so that when the user selects Countries it would display the result from the query alphabatically and if he selects Numbers then list based on numbers in descending.

Having researched, I have applied this (echo concept into my code) but doesn't seem to work.

I have the following query written in PHP that output staffs' country of birth( no of staff born in number of countries) in ascending order:

$querytest = "select x , COUNT( * )  from( select `staffbirthplace` as x from staffbirthdetails where staffemailid IN(SELECT staffemailid FROM staff where orgid='" . $orgId . "'  AND deptname='" . $deptName . "' AND teamname='" . $teamName . "') ) as temptable group by x order by count(*) ASC ";

Then, I have a dropdown in HTML:

<form>
    <label for="sortorder">Sort by:</label>
    <select id="sortByDropdown" onchange="sortBy(this);">
      <option value="alphabatically">Countries</option>
      <option value="numbers">Number</option>
    </select>
</form>

Furthermore, I have a Javascript function sortBy()

function sortBy(){
    var sortByDrpdownDiv =  document.getElementById('sortByDropdown');

if (sortByDrpdownDiv[sortByDrpdownDiv.selectedIndex].value == 'numbers'){
    alert("yo in if statement");
    <?php $querytest = "select x , COUNT( * ) from( select `staffbirthplace` as x from staffbirthdetails where staffemailid IN(SELECT staffemailid FROM staff where orgid='" . $orgId . "'  AND deptname='" . $deptName . "' AND teamname='" . $teamName . "') ) as temptable group by x order by count(*) DESC ";
    $result = mysql_query($querytest);

    while ($row = mysql_fetch_assoc($result)) {
            echo "<b>";
            echo $row['x'];
            echo ": </b>&nbsp;";
            echo $row['COUNT( * )'];
            echo "<br/>";
        }?>
    document.getElementById('staffbirthplaces').innerHTML = <?php echo $row?>;
    }
}

First I am going only for Numbersbecause the same logic will apply to the Countries. Any help will be appreciated

Answer 1

So, i finally did it! Used switch instead of IF statement. Below is the code:

   function sortByAlphabetsOrNumbers(obj){

    var selectedValue = obj.options[obj.selectedIndex].value
switch(selectedValue)
{
    case "numberOfStaff":
        document.getElementById('sortBy').innerHTML = 
        "<?php
            include 'connection.php';

            $staffNumbersDesc = "select x , COUNT( * )  from( select `staffbirthplace` as x from staffbirthdetails where staffemailid IN(SELECT staffemailid FROM staff where orgid='" . $orgId . "'  AND deptname='" . $deptName . "' AND teamname='" . $teamName . "') ) as temptable group by x order by count(*) DESC";
            $result = mysql_query($staffNumbersDesc);               
            while ($row = mysql_fetch_assoc($result)) 
            {
                echo "<b>";
                echo $row['x'];
                echo ": </b>&nbsp;";
                echo $row['COUNT( * )'];
                echo "<br/>";
                } 
        ?>";
        document.getElementById('birthCountriesAlphabaticalOrder').style.display = "none";
        break;

    case "countries":
        document.getElementById('sortBy').innerHTML = 
        "<?php
            include 'connection.php';

            $alphabaticalOrder = "select x , COUNT( * )  from( select `staffbirthplace` as x from staffbirthdetails where staffemailid IN(SELECT staffemailid FROM staff where orgid='" . $orgId . "'  AND deptname='" . $deptName . "' AND teamname='" . $teamName . "') ) as temptable group by x";
            $result = mysql_query($alphabaticalOrder);              
            while ($row = mysql_fetch_assoc($result)) 
            {
                echo "<b>";
                echo $row['x'];
                echo ": </b>&nbsp;";
                echo $row['COUNT( * )'];
                echo "<br/>";
                } 
        ?>";
        document.getElementById('birthCountriesAlphabaticalOrder').style.display = "none";
        break;
}
};

Hope it helps someone