Php 如何在html中填充选项根据用户填充的html字段选择。
下面是代码片段:Php 如何在html中填充选项根据用户填充的html字段选择。,php,mysqli,Php,Mysqli,下面是代码片段: <select name="isbn" onchange=" "> <?php //Displaying all ISBN in drop down $result = mysqli_query($con,"SELECT * FROM book"); while($row = mysqli_fetch_array($result)) { ?> <option> <?php echo $row['ISBN']; ?>&l
<select name="isbn" onchange=" ">
<?php
//Displaying all ISBN in drop down
$result = mysqli_query($con,"SELECT * FROM book");
while($row = mysqli_fetch_array($result)) {
?>
<option> <?php echo $row['ISBN']; ?></option>
<?php } ?>
</select>
</label></td>
</tr>
<tr>
<td>Copy Number </td>
<td><select name="copy_number">
<?php
//Now based on selected book I want to fetch number of copies from database
$result = mysqli_query($con,"SELECT number_of_copies FROM book where isbn = [ SELECTE VALUE FROM ABOVE]");
?>
副本号
我该怎么做呢?您可以使用简单的php处理程序来获取所选书籍的副本
<select name="isbn">
.....
</select>
<div id="copies"></div>
<script>
$.ajax({
url: "getCopies.php?",
data: "isbn=" + $("select[name='isbn']").val(),
type: "POST",
dataType: "json",
success: function(response) {
$.each(response, function(i, item) {
$("#copies").append(item.name); // Sample json format {id: "213123", name:"Lord of the rings", isbn:"887799..."}
})
}
});
</script>
.....
$.ajax({
url:“getCopies.php?”,
数据:“isbn=“+$”(“选择[name='isbn']”)val(),
类型:“POST”,
数据类型:“json”,
成功:功能(响应){
$。每个(响应、功能(i、项目){
$(“#copies”).append(item.name);//示例json格式{id:“213123”,名称:“指环王”,isbn:“887799…”
})
}
});
getCopies.php
<?php
$isbn = $_POST["isbn"];
// Some db connections
$result = mysqli_query($con,"SELECT number_of_copies FROM book where isbn = $isbn");
$resultArr = array();
while($row = mysqli_fetch_array($result)) {
$resultArr[] = $row;
}
echo json_encode($resultArr); // This will return rows in json format. You can iterate it in js side
您需要使用ajax来填充预先填充的选项,或者在第一个选择更改后强制页面刷新,并使用其值填充第二个下拉列表。如果SQL查询返回单个值,为什么它需要位于选择
控件中?您可以使用ajax和sipmle php处理程序来获取所选选项的副本。更多细节见我的答案