我想在输入字段旁边显示php错误消息。我可以这样做吗?
这是我在同一页面上的php和html代码。它只是简单的测试代码..任何帮助都会很棒
尝试将php放在输入旁边,但这不起作用..我把它放在一个值
中<?php
if ( isset( $_POST['submit'] ) ) {
$mysqli = new mysqli("localhost", "root", "", "test");
$fname = $_POST["fname"];
$lname = $_POST["lname"];
if (empty($henny)) { $errorhenny = '<div class="isa_error">Frist name is required<br></div>'; }
if (empty($lname)) {
echo '<div class="isa_error">Last name is required<br></div>';
}
else{
$stmt = $mysqli->prepare("INSERT INTO test (Firstname, Lastname) VALUES (?, ?)");
$stmt->bind_param("ss", $fname, $lname);
$stmt->execute();
$stmt->close();
$mysqli->close();
}
}
?>
<html>
<body>
<form action="" method="post">
<input type="text" name="fname" value= "<?php echo $errorhenny; ?>">
<input type="text" name="lname">
<input type="submit" name="submit" value="Submit">
</form>
</html>
答案 0 :(得分:0)
尝试使用ajax ...更容易和更好
答案 1 :(得分:0)
在php变量
中分配错误消息if (empty($fname)) {
$errorfname= '<div class="isa_error">Frist name is required</div>';
}
if (empty($lname)) {
$errorlname = '<div class="isa_error">Last name is required</div>';
}
然后在html中打印变量
<html>
<body>
<form action="" method="post">
<input type="text" name="fname">
<?php if(isset($errorfname)) { echo $errorfname; } ?> // fname error message
<input type="text" name="lname">
<?php if(isset($errorlname)) { echo $errorlname; } ?> //lname error message
<input type="submit" name="submit" value="Submit">
</form>
答案 2 :(得分:0)
试试这个:
<?php
$ferr = '';
$lerr = '';
if ( isset( $_POST['submit'] ) ) {
$mysqli = new mysqli("localhost", "root", "", "test");
$fname = $_POST["fname"];
$lname = $_POST["lname"];
if (empty($fname)) {
$ferr = 'First name is required';
}
if (empty($lname)) {
$lerr = 'Last name is required';
}
else{
$stmt = $mysqli->prepare("INSERT INTO test (Firstname, Lastname) VALUES (?, ?)");
$stmt->bind_param("ss", $fname, $lname);
$stmt->execute();
$stmt->close();
$mysqli->close();
}
}
?>
<html>
<body>
<form action="" method="post">
<div>
<input type="text" name="fname">
<span class="isa_error"><?php echo $ferr;?></span>
</div>
<div>
<input type="text" name="lname">
<span class="isa_error"><?php echo $lerr;?></span>
</div>
<input type="submit" name="submit" value="Submit">
</form>
</html>
答案 3 :(得分:0)
<?php
if ( isset( $_POST['submit'] ) ) {
$mysqli = new mysqli("localhost", "root", "", "test");
$fname = $_POST["fname"];
$lname = $_POST["lname"];
if(empty($fname) || empty($lname)){
if (empty($fname)) {
$errorfname = '<div class="isa_error">Frist name is required<br></div>';
}
if (empty($lname)) {
$errorlastname = '<div class="isa_error">Last name is required<br></div>';
}
}
else{
$stmt = $mysqli->prepare("INSERT INTO test (Firstname, Lastname) VALUES (?, ?)");
$stmt->bind_param("ss", $fname, $lname);
$stmt->execute();
$stmt->close();
$mysqli->close();
}
}
?>
<html>
<body>
<form action="" method="post">
<input type="text" name="fname" value= ""><?php echo isset($errorfname) ? $errorfname : ''; ?>
<input type="text" name="lname"><?php echo isset($errorlastname) ? $errorlastname : ''; ?>
<input type="submit" name="submit" value="Submit">
</form>
</html>