Question

非常新的PHP。如果这个问题很荒谬，请原谅我。无论如何，我有一些PHP将数据发送到数据库。以下是其中一些内容：

else {

$sql = "INSERT INTO users (username, firstname, lastname, password)
VALUES ('$username', '$firstname', '$lastname', '$pass')";

$sql = "CREATE TABLE $username (
day VARCHAR(10) NOT NULL, 
P1 VARCHAR(10) NOT NULL,
P1_WORK VARCHAR(30) NOT NULL,
P2 VARCHAR(10) NOT NULL,
P2_WORK VARCHAR(30) NOT NULL,
P3 VARCHAR(10) NOT NULL,
P3_WORK VARCHAR(30) NOT NULL,
P4 VARCHAR(10) NOT NULL,
P4_WORK VARCHAR(30) NOT NULL,
P5 VARCHAR(10) NOT NULL,
P5_WORK VARCHAR(30) NOT NULL,
P6 VARCHAR(10) NOT NULL,
P6_WORK VARCHAR(30) NOT NULL
)";

if (mysqli_query($con, $sql)) {
    echo "okay";
} else {
    echo "error";
}
    }

当我运行它时，我意识到只会执行第二个$sql=，因为我在同一个实例中定义了$sql两次。我怎样才能这样做，所以这两个事件都是一次性触发的？谢谢！

编辑：完整代码

<?php
echo error_reporting(E_ALL);
$username = strval($_GET['enteredUserName']);
$pass = strval($_GET['enteredPassword']);
$firstname = strval($_GET['enteredFirstName']);
$lastname = strval($_GET['enteredLastName']);

$con = mysqli_connect('localhost','XXXX','XXXX','XXXX');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
} else {
$sql="SELECT * FROM users WHERE username = '".$username."'";
$result = mysqli_query($con,$sql);
if ($result->num_rows > 0)  {
    echo "pop";
} else {

$sql = "INSERT INTO users (username, firstname, lastname, password)
VALUES ('$username', '$firstname', '$lastname', '$pass') ";

$sql = "CREATE TABLE $username (
day VARCHAR(10) NOT NULL, 
P1 VARCHAR(10) NOT NULL,
P1_WORK VARCHAR(30) NOT NULL,
P2 VARCHAR(10) NOT NULL,
P2_WORK VARCHAR(30) NOT NULL,
P3 VARCHAR(10) NOT NULL,
P3_WORK VARCHAR(30) NOT NULL,
P4 VARCHAR(10) NOT NULL,
P4_WORK VARCHAR(30) NOT NULL,
P5 VARCHAR(10) NOT NULL,
P5_WORK VARCHAR(30) NOT NULL,
P6 VARCHAR(10) NOT NULL,
P6_WORK VARCHAR(30) NOT NULL
)";

if (mysqli_query($con, $sql)) {
    echo "okay";
} else {
    echo "error";
}
    }
mysqli_close($con);
}
?>

Answer 1

就像u_mulder在评论中所说，你可以这样做，就像mysqli_error returns empty string if there is no error occured一样。

$sql1 = 'first query';
$sql2 = 'second query';
mysqli_query($h, $sql1);
mysqli_query($h, $sql2);
$errorOccured = false;
if (strlen(mysqli_error($h)) > 1) {
    $errorOccured = true;
}

查看mysqli_error in the manual。

或者，正如已经指出的那样，只需执行两个mysqli_queries并检查它们的返回值。或者，如果您需要整体执行查询，您还可以使用名为＆＃34; transactions＆＃34;的内容，查看其中一个注释示例。取自mysqli_rollback in the manual：

    $all_query_ok=true; // our control variable 

//we make 4 inserts, the last one generates an error 
//if at least one query returns an error we change our control variable 
$mysqli->query("INSERT INTO myCity (id) VALUES (100)") ? null : $all_query_ok=false; 
$mysqli->query("INSERT INTO myCity (id) VALUES (200)") ? null : $all_query_ok=false; 
$mysqli->query("INSERT INTO myCity (id) VALUES (300)") ? null : $all_query_ok=false; 
$mysqli->query("INSERT INTO myCity (id) VALUES (100)") ? null : $all_query_ok=false; //duplicated PRIMARY KEY VALUE 

//now let's test our control variable
// This is shorthand if else statement
// ($statement) ? If true : if false;
$all_query_ok ? $mysqli->commit() : $mysqli->rollback(); 

$mysqli->close();

Answer 2

为您的两个查询使用$sql1和$sql2两个不同的变量，然后

要么（一次两个）： -

if(mysqli_query($con, $sql1) && mysqli_query($con, $sql2)){
   echo "Ok";
}else{
   echo "Error";
}

或逐一： -

if (mysqli_query($con, $sql)) {
    if(mysqli_query($con, $sql2)){
        echo "OK";
     }else{
        echo "error";
     }
} else {
    echo "error";
}

或者您需要检查mysqli_multi_query()： -

http://php.net/manual/en/mysqli.multi-query.php

Answer 3

为不同的查询使用不同的变量

$sql1 = "INSERT INTO users (username, firstname, lastname, password)
VALUES ('$username', '$firstname', '$lastname', '$pass')";

$sql2 = "CREATE TABLE $username (
day VARCHAR(10) NOT NULL, 
P1 VARCHAR(10) NOT NULL,
P1_WORK VARCHAR(30) NOT NULL,
P2 VARCHAR(10) NOT NULL,
P2_WORK VARCHAR(30) NOT NULL,
P3 VARCHAR(10) NOT NULL,
P3_WORK VARCHAR(30) NOT NULL,
P4 VARCHAR(10) NOT NULL,
P4_WORK VARCHAR(30) NOT NULL,
P5 VARCHAR(10) NOT NULL,
P5_WORK VARCHAR(30) NOT NULL,
P6 VARCHAR(10) NOT NULL,
P6_WORK VARCHAR(30) NOT NULL
)";

if (mysqli_query($con, $sql1) && mysqli_query($con, $sql2)) {
    echo "okay";
} else {
    echo "error";
}

如何触发这两个？

3 个答案: