如何將SQL語言與PHP語言混合使用

Question

我寫了這段代碼：

<?php
$con = mysqli_connect('localhost', 'root', '');
if(!$con)
{
    die("not ok");
}

mysqli_select_db($con,"uoh");  

$q = " SELECT * FROM student WHERE id = 201102887" ;
$result = mysqli_query($con , $q ) ;
if($row = mysqli_fetch_array($result))
{
   echo "<h3> compliance for for " . $row["name"];
   echo " and the major is ".$row["major"];
   echo "</h3>";
}

$major=$row["major"];

$con = mysqli_connect('localhost', 'root', '');
if(!$con)
{
    die("not ok");
}

mysqli_select_db($con,"uoh");  
$q = "SELECT * FROM courses LEFT JOIN equal ON equal.course_number= 
courses.course_number LEFT JOIN degree_plan ON degree_plan.course_number=
 courses.course_number LEFT JOIN student_record ON courses.course_number= 
 student_record.course_number AND student_record.id=201102887 AND degree_plan.major=".$major;
?>

該代碼工作正常，但沒有給我結果。

我認為問題出在degree_plan.major=".$major;因為查詢沒有給我結果。

你能解決嗎？

Answer 1

您不會執行僅定義為字符串的第二個查詢。 您應該在第二個查詢之后放置與第一個查詢相同的內容

$result2 = mysqli_query($con , $q ) ;
if($row2 = mysqli_fetch_array($result2)) {

由於主值看起來像一個字符串，您可能需要用引號引起來，否則查詢將失敗

AND degree_plan.major='".$major."'";

Answer 2

您應該這樣用引號將查詢格式化

    mysqli_select_db($con,"uoh");  
    $q = "SELECT * FROM courses LEFT JOIN equal ON equal.course_number= 
     courses.course_number LEFT JOIN degree_plan ON degree_plan.course_number=
     courses.course_number LEFT JOIN student_record ON courses.course_number= 
     student_record.course_number 
      AND student_record.id=201102887 AND degree_plan.major='".$major ."'";