这是我的编码。当我单击表上的任何行时,我想查看数据。但是它仅查看行中的第一个数据。您认为出了什么问题?
我有关从数据库检索数据的代码正在运行。但是,当我单击它时,它总是查看第一个检索到的数据行的数据。
<?php
$sql = "SELECT * FROM courses";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table class='table table-bordered' id='myTable'>";
echo "<tr>";
echo "<th>COURSE CODE</th>";
echo "<th>COURSE NAME</th>";
echo "<th>ACTION</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['course_code'] . "</td>";
echo "<td>" . $row['course_desc'] . "</td>";
echo "<td>";
echo "<a href='#' data-toggle='modal' data-target='#confirmenrollment'>";
echo "Enroll";
echo "</a>";
echo "</td>";
echo "</tr>";
echo "<div class='modal fade' id='confirmenrollment' tabindex='-1' role='dialog'>";
echo "<div class='modal-dialog'>";
echo "<div class='modal-content'>";
echo "<div class='modal-header'>";
echo "<button type='button' class='close' data-dismiss='modal'>×</button>";
echo "<h3>Confirm course to enroll.</h3><br>";
echo "<p><b>Course Name:   </b>" . $row['course_desc'] . "</p>";
echo "<p><b>Course Code:   </b>" . $row['course_code'] . "</p>";
echo "<p><b>Price:   </b>" . $row['course_price'] . "</p>";
echo "<div class='more-button'>";
echo "<a href='#'>";
echo "Confirm Enrollment";
echo "</a>";
echo "</div>";
echo "</div>";
echo "</div>";
echo "</div>";
echo "</div>";
}
echo "</table>";
mysqli_free_result($result);
} else{
echo "No courses found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
mysqli_close($link);
?>
谁可以帮我这个事?
您的问题是所有模态都具有相同的DOM ID。您需要它们是唯一的,并且模态的目标必须反映这一点,以便每一行都可以打开正确的模态。
假设它course_code
在数据库中是唯一的,则可以使用它。然后我们更新该行以打开模态,
echo "<a href='#' data-toggle='modal' data-target='#confirmenrollment-".$row['course_code']."'>";
还有模式ID本身
echo "<div class='modal fade' id='confirmenrollment-".$row['course_code']."' tabindex='-1' role='dialog'>";
请注意,我是如何附加-".$row['course_code']."
到data-target
属性和模式ID的。
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我来说两句