下拉列表是从数据库“meal_planner”中填充的使用第一个选择。我只是把它拼凑在一起。
<?php
$link=mysqli_connect("localhost", "root", "");
mysqli_select_db($link,"meal_planner");
?>
<!DOCTYPE html>
<html>
<head>
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
};
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send();
}
}
</script>
</head>
<body>
<form name="form1" action="" method="post">
<select name="users" onchange="showUser(this.value)">
<option value="">Select a person:</option>
<?php
$res=mysqli_query($link,"select name from protein");
while($row=mysqli_fetch_array($res))
{
?>
<option><?php echo $row["name"]; ?></option>
<?php
}
?>
</select>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here...</b></div>
</body>
</html>
获取用户.php
<!DOCTYPE html>
<html>
<head>
<style>
table {
width: 100%;
border-collapse: collapse;
}
table, td, th {
border: 1px solid black;
padding: 5px;
}
th {text-align: left;}
</style>
</head>
<body>
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost','root','','meal_planner');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM protein WHERE protein_id = '".$q."'";
$result = mysqli_query($con,$sql);
echo "<table>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Job</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['calories'] . "</td>";
echo "<td>" . $row['protein'] . "</td>";
echo "<td>" . $row['carbs'] . "</td>";
echo "<td>" . $row['fats'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
</html>
最佳答案
您没有输入 protein_id进入<option>元素。
$res=mysqli_query($link,"select name, protein_id from protein");
while($row=mysqli_fetch_array($res))
{
?>
<option value="<?php echo $row["protein_id"]; ?>"><?php echo $row["name"]; ?></option>
<?php
}
当没有 value 时属性，它使用元素的文本，因此您将蛋白质名称作为选择值结束。然后当你使用 intval($q) , 它返回 0因为蛋白质名称不是有效整数，并且 AJAX 调用中的查询总是使用 protein_id = 0 获取蛋白质.
关于php - 如果输入更改，用户填充框将不会更改，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39627756/