使用php函数对数字索引数组的元素进行分级

问题描述：

I intend to grade the elements of a numerically indexed array in php using function call; my function call does not work: it doesn't grade the elements and I cannot figure out the error in the code.Please help Thanks, in advance

 <?php

//function intended to  grade array elements

    function gradeArray($x){
        if($score>= 70){
    echo"A";


     }


       elseif($score >= 50){


     echo"B";
    }
    elseif($score>= 40){


     echo"C";


     }


     else{
    echo"F";


     }
    }

    // Array of Scores to be graded


      $scores = array ("55", "68", "43", "78");
//Display result in a tabular form


     echo"<table  border = '1'><th>Score</th><th>Grade</th>";

    foreach($scores as $score){



     echo"<tr><td>";



      echo$score."</td>";

    echo"<td>". gradeArray($score);

    echo"</td></tr>";



      }



       echo"</table>";

    ?>

我打算使用函数调用在php中对数字索引数组的元素进行分级; 我的函数调用不起作用：它没有对元素进行分级，我无法弄清楚代码中的错误。请帮助谢谢，提前 p>

 ＆lt;？php  
 
 //用于对数组元素进行分级的函数
 
函数gradeArray（$ x）{
 if（$ score＆gt; = 70）{
 echo“A”; 
 
 
} 
  
 
 elseif（$ score＆gt; = 50）{
 
 
 echo“B”; 
} 
 elseif（$ score＆gt; = 40）{
 
 
 echo“C”;  
 
 
} 
 
 
其他{
 echo“F”; 
 
 
} 
} 
 
 //要评分的分数数组
 
 
  $ scores = array（“55”，“68”，“43”，“78”）; 
 //以表格形式显示结果
 
 
 echo“＆lt; table border ='1'＆gt;  ＆lt; th＆gt;得分＆lt; / th＆gt;＆lt; th＆gt;等级＆lt; / th＆gt;“; 
 
 foreach（$得分为$得分）{
 
 
 
回声”＆lt; tr＆gt;＆lt; td＆gt;  ;“; 
 
 
 
 echo $ score。”＆lt; / td＆gt;“; 
 
 echo”＆lt; td＆gt;“。  gradeArray（$ score）; 
 
 echo“＆lt; / td＆gt;＆lt; / tr＆gt;”; 
 
 
 
} 
 
 
 
 echo“＆lt; / table＆gt;”;  
 
？＆gt; 
  code>  pre> 
  div>

答

You are passing $x into your function then calling $score. Your scores array is also in string format, just need to remove the quotes to make them numbers. Also change $x to $score and it should work fine! :)

<?php
//function intended to  grade array elements
function gradeArray($score) {
   if     ($score >= 70)  return "A";
   elseif ($score >= 50)  return "B";
   elseif ($score >= 40)  return "C";
   else                   return "F";
}

// Array of Scores to be graded
$scores = array (55, 68, 43, 78);

//Display result in a tabular form
echo "<table border='1'><th>Score</th><th>Grade</th>";

foreach ($scores as $score) {
    echo "<tr><td>$score</td><td>" . gradeArray($score) . "</td></tr>";
}

echo "</table>";
?>

答

Your array elements are in string .convert all element to int using

    gradeArray($x){
$score=(int)$x;
}

Try this it will work

答

First off, most likely $score is undefined in your function since you use $x.

function gradeArray($x){

Then you are using your if conditions as if($score>= 70){.

Also, in your return values, just use return.

return"A"; // and others

Use return not echo so that this concatenation echo "<td>". gradeArray($score); works.

答

$score is undefined in your variable, the function is fetching the variable as $x try to change $score to $x or define $score., you can use global $score also in your function. Also you should return the values instead of using echo use the code below

<?php

//function intended to  grade array elements

    function gradeArray($x){
$score=$x;
        if($score>= 70){
    return "A";


     }


       elseif($score >= 50){


     return "B";
    }
    elseif($score>= 40){


    return "C";


     }


     else{
    return "F";


     }
    }

    // Array of Scores to be graded


      $scores = array ("55", "68", "43", "78");
//Display result in a tabular form


     echo"<table  border = '1'><th>Score</th><th>Grade</th>";

    foreach($scores as $score){



     echo"<tr><td>";



      echo$score."</td>";

    echo"<td>". gradeArray($score);

    echo"</td></tr>";



      }



       echo"</table>";

    ?>

Hope this helps you

答

Try this

<?php

//function intended to  grade array elements

function gradeArray($x) {
    if ($x >= 70) {
        return "A";
    } elseif ($x >= 50) {

        return "B";
    } elseif ($x >= 40) {

        return "C";
    } else {
        return "F";
    }
}

// Array of Scores to be graded


$scores = array("55", "68", "43", "78");
//Display result in a tabular form


echo"<table  border = '1'><th>Score</th><th>Grade</th>";

foreach ($scores as $score) {



    echo"<tr><td>";



    echo$score . "</td>";

    echo"<td>" . gradeArray($score);

    echo"</td></tr>";
}



echo"</table>";
?>

答

function grading( $marks ){
    $grade = mysql_query("SELECT grade_name,grade_point FROM table_name **strong text**WHERE smark <= round($marks) AND hmark >= round($marks)");

     $gd = mysql_fetch_row( $grade );

     return $gd[0];
}

使用php函数对数字索引数组的元素进行分级

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