Question

我在mysql中有几个单独的表，名为tblstudent和tblskill。 tblskill根据studentId保存数据。现在的问题是当我想显示数据时，显示的数据将被复制，具体取决于ID持有的技能。知道如何解决这个问题吗？这是我的数据显示的屏幕截图screenshot这是我的tblskill

attributes[skillId(int),skill(varchar),studentId(varchar)]

这是我的代码：

<?php 

$result = $mysqli->query("SELECT tblstudent.id,tblstudent.studentId,tblstudent.programme,tblstudent.cgpa,tblpersonalinfo.studentId,tblpersonalinfo.pImage,tblpersonalinfo.pImageType,tblpersonalinfo.pImageSize,tblpersonalinfo.pName FROM tblstudent INNER JOIN tblpersonalinfo ON tblstudent.studentId = tblpersonalinfo.studentId ".$where_sql."  ");

  if ($result->num_rows != 0) {



  echo "<table class='scroll' width='700' border='0' bgcolor='#FF00FF' align='left'>";
  echo "<tbody>";



 while($row = mysqli_fetch_array($result)) {

  $studentId = $row[1];


 // get data from db


  echo "<input type='hidden' name='id' value='".$row[0]."'>";
  echo "<tr>";
  echo "<td>";
  echo '<div  class="title bg-primary" align="left" style="font-weight: bold;">' .   '&nbsp;' .  '<img src="uploads/' . $row['pImage'] .' " alt = "avatar" height="60" width="50" >' . '&nbsp;' . $row['pName'] . '&nbsp;' . '&nbsp;' . '&nbsp;' . '<a href="viewResume.php?id=' .  $row[0]  . ' " target="_blank" >'.'<img src="image/viewResume.png" alt="Apply Now" width="100" height="40" border="0">'.'</a>'. '</div>';
  echo '<div  align="left" class="sub">' . '<br>' . $row['programme']. '</div>';





  include('Connections/connect.php');


    if ($result1 = $mysqli->query("SELECT * FROM tblskill WHERE studentId=$studentId"))
    {
  if ($result1->num_rows != 0) {
    while($row = mysqli_fetch_array($result1)) {

      echo "<table class='scroll' width='700' border='0' bgcolor='#FF00FF' align='left'>";
     echo "<tbody>";
     echo "<tr>";
     echo "<td>";
     echo '<div  align="left" class="sub">' . '<br>' . $row['skill']. '</div>';
     echo "</td>";
     echo "</tr>";
     echo "</tbody>";
     echo "</table>";

}
}
}

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

}
echo "<tr>";
echo '<td>' . '&nbsp;' . '</td>';
echo "</tr>";
echo "</tbody>";
echo "</table>";

} 
?>

Answer 1

$result = $mysqli->query("SELECT DISTINCT tblstudent.id,tblstudent.studentId,tblstudent.programme,tblstudent.cgpa,tblpersonalinfo.studentId,tblpersonalinfo.pImage,tblpersonalinfo.pImageType,tblpersonalinfo.pImageSize,tblpersonalinfo.pName FROM tblstudent INNER JOIN tblpersonalinfo ON tblstudent.studentId = tblpersonalinfo.studentId ".$where_sql."  ");

在SELECT之后添加到您的查询DISTINCT http://www.w3schools.com/sql/sql_distinct.asp

Mysqli查询里面的查询

1 个答案: