我在尝试将用户输入的数据输出到表单中时遇到问题。这是一个2块表单,我只是试图将数据保存到2个变量中,然后回显变量。但我不明白我哪里错了。感谢您的帮助。
<?php
echo $problem = "";
if(isset($_POST['submit']) && $_POST['submit']=="submit"){
if(!empty($_POST['eWeight']) && $_POST['eWeight']!=''){
$eWeight = mysqli_real_escape_string($conn, trim($_POST['eWeight']));
} else {
$problem .= "Please enter a weight. <br/>";
}
if(!empty($_POST['gym']) && $_POST['gym']!=''){
$gym = mysqli_real_escape_string($conn, trim($_POST['gym']));
} else {
$problem .= "Please enter time at gym. <br/>";
}
echo $eWeight, $gym;
}
?>
<?php
if($conn){
echo "connected";
}
echo $problem;
echo $eWeight;
?>
<form class="pure-form pure-form-stacked" name="contact_weight">
<label for="eWeight">Enter Weight: </label> <input type="number" id="eWeight" name="eWeight" placeholder="88" required/>
<label for="gym">Enter Time at Gym: </label> <input type="number" id="gym" name="gym" placeholder="60" required/>
<button class="submit" type="submit">Submit Form</button>
</form>
您可以执行类似的操作
<?php
echo $problem = "";
if(isset($_POST['submit'])){
if(!empty($_POST['eWeight']) && $_POST['eWeight']!=''){
$eWeight = mysqli_real_escape_string($conn, trim($_POST['eWeight']));
} else {
$problem .= "Please enter a weight. <br/>";
}
if(!empty($_POST['gym']) && $_POST['gym']!=''){
$gym = mysqli_real_escape_string($conn, trim($_POST['gym']));
} else {
$problem .= "Please enter time at gym. <br/>";
}
echo $eWeight, $gym;
}
if($conn){
echo "connected";
}
echo $problem;
echo $eWeight;
?>
<form class="pure-form pure-form-stacked" name="contact_weight" action="" method="post">
<label for="eWeight">Enter Weight: </label> <input type="number" id="eWeight" name="eWeight" placeholder="88" required/>
<label for="gym">Enter Time at Gym: </label> <input type="number" id="gym" name="gym" placeholder="60" required/>
<button name="submit" class="submit" type="submit">Submit Form</button>
</form>