让我开始说我是php的新手,所以我仍然在努力学习。我在Wordpress上创建了一个表单,我想在一个表上插入值(data_test表,我已经管理过),然后从data_test表中获取所有列(id是自动增量号,名称,电子邮件,产品,数量用户输入)并插入到其他表。我使用这个html代码来表单来解析值:
<form action="../enter_data_insert.php" method="post" onsubmit="return form_validation()" name="myForm">
Name <input id="name" name="name" type="text" />
Email <input id="email" name="email" required type="email"/>
Product<input id="prod" name="prod" required type="text" />
Quantity<input id="quant" name="quant" required type="number" min="1" / >
<input type="submit" value="Submit" />
</form>
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然后这个php取值:
<?php
if(!empty($_POST["name"]) && !empty($_POST["email"]) && !empty($_POST["prod"]) && !empty($_POST["quant"])){
//connect with database
include "database_conn.php";
//get the form elements and store them in variables
session_start();
$name=$_POST["name"];
$email=$_POST["email"];
$prod=$_POST["prod"];
$quant=$_POST["quant"];
//insert data on data_test table
$sql="INSERT INTO `site_db`.`data_test` ( `name` , `email`, `prod`,`quant`) VALUES ( '$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//retrieve data
$sql = "SELECT data_test_id FROM data_test WHERE prod='$prod'";
$result = mysqli_query($con,$sql);
if(!$result){
echo mysqli_error($con);
} else{
while($value = mysqli_fetch_object($result)){
$id = intval($value->id);
$_SESSION['myid'] = $value->id;
var_dump($value);
//insert data on data_test_ins table
$sql="INSERT INTO site_db.data_test_ins` ( id,name , email, prod,quant) VALUES ( $id,'$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//Redirects to the specified page
// header("Location: http://localhost/site/");
}
}
}
}
}
?>
&#13;
现在它在data_test表上插入除id之外的所有值,我猜它是null,因为它必须关闭php上的第一个插入,然后我必须调用第二个插入(//在data_test_ins表上插入数据)在其他PHP? 但我不确定,有人能帮帮我吗?或者只是指导我做正确的方法。 我开始认为我必须创建两个PHP来解析值并采取第一个表,然后在另一个PHP上插入值? 任何想法都有帮助! : - )
答案 0 :(得分:1)
你在做什么是不对的。手动向数据库添加id字段的值不是一个好方法。它应该由数据库自动生成。我建议的是,在data_test_ins表中添加另一个字段,例如:test_id,它指向data_test表的id。这是外键的概念。 阅读有关外键here
的概念您的代码现在是: -
<?php
if(!empty($_POST["name"]) && !empty($_POST["email"]) && !empty($_POST["prod"]) && !empty($_POST["quant"])){
//connect with database
include "database_conn.php";
//get the form elements and store them in variables
session_start();
$name=$_POST["name"];
$email=$_POST["email"];
$prod=$_POST["prod"];
$quant=$_POST["quant"];
//insert data on data_test table
$sql="INSERT INTO `site_db`.`data_test` ( `name` , `email`, `prod`,`quant`) VALUES ( '$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//retrieve data
$sql = "SELECT data_test_id FROM data_test WHERE prod='$prod'";
$result = mysqli_query($con,$sql);
if(!$result){
echo mysqli_error($con);
} else{
while($value = mysqli_fetch_object($result)){
$id = $value->id;
//insert data on data_test_ins table
$sql="INSERT INTO `site_db`.`data_test_ins` ( `id`,`name` , `email`, `prod`,`quant`, `test_id`) VALUES ('$name','$email','$prod','$quant', '$id')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//Redirects to the specified page
header("Location: http://localhost/site/");
}
}
}
}
}
?>
答案 1 :(得分:0)
您是否尝试将$value->id
传递给查询而不是$value
?
它是一个具有结果集当前行的对象,因此您只应传递此对象的id属性。
$sql="INSERT INTO `site_db`.`data_test_ins` ( `id`,`name` , `email`, `prod`,`quant`) VALUES ( '$value->id','$name','$email','$prod','$quant')";
增加:
mysql
已弃用的库。 编辑:
您的代码应如下所示:
<?php
if(!empty($_POST["name"]) && !empty($_POST["email"]) && !empty($_POST["prod"]) && !empty($_POST["quant"])){
//connect with database
include "database_conn.php";
//get the form elements and store them in variables
session_start();
$name=$_POST["name"];
$email=$_POST["email"];
$prod=$_POST["prod"];
$quant=$_POST["quant"];
//insert data on data_test table
$sql="INSERT INTO `site_db`.`data_test` ( `name` , `email`, `prod`,`quant`) VALUES ( '$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//retrieve data
$sql = "SELECT data_test_id FROM data_test WHERE prod='$prod'";
$result = mysqli_query($con,$sql);
if(!$result){
echo mysqli_error($con);
} else{
while($value = mysqli_fetch_object($result)){
$_SESSION['myid'] = $value->data_test_id;
$id = intval($value->data_test_id);
//insert data on data_test_ins table
$sql="INSERT INTO `site_db`.`data_test_ins` ( id,name , email, prod,quant) VALUES ( '$id','$name','$email','$prod','$quant')";
if(!mysqli_query($con,$sql)){
echo mysqli_error($con);
} else{
//Redirects to the specified page
header("Location: http://localhost/site/");
}
}
}
}
}
?>