PHP代码在PHP不行CLI

我与数据库连接的PHP代码名为test.php的象下面这样：PHP代码在PHP不行CLI

<? 
$mysqli = new mysqli("localhost","root","","monster"); 

/* check connection */ 
if (mysqli_connect_errno()) { 
    printf("Connect failed: %s\n", mysqli_connect_error()); 
    exit(); 
} 

/* return name of current default database */ 
if ($result = $mysqli->query("INSERT INTO `earth` (`monster_id`, `type`, `name`) VALUES 
(NULL, 'defender', 'tortoise') 
")) { 
    $row = $result->fetch_row(); 
    printf("Default database is %s.\n", $row[0]); 
    $result->close(); 
} 
$mysqli->close(); 
?>

它工作时，我通过插入到我的数据库浏览器http://localhost/learn/test.php和数据呼叫

但当我运行在PHP CLI

php -r ' $mysqli = new mysqli("localhost","root","","monster"); 

    /* check connection */ 
    if (mysqli_connect_errno()) { 
     printf("Connect failed: %s\n", mysqli_connect_error()); 
     exit(); 
    } 

    /* return name of current default database */ 
    if ($result = $mysqli->query("INSERT INTO `earth` (`monster_id`, `type`, `name`) VALUES 
    (NULL, 'defender', 'tortoise') 
    ")) { 
     $row = $result->fetch_row(); 
     printf("Default database is %s.\n", $row[0]); 
     $result->close(); 
    } 
    $mysqli->close();'

没有插入数据。更新现在我试用eval（$ code）;似乎OKAY在Windows中，但不是在LINUX

来源

2014-01-29 Ronnie Aban

什么样的错误你得到些什么？ –

语法高亮显示似乎认为你有一个报价逃跑问题.. – Ben

现在我试用eval（$ code）; （NULL，\'defender \'，\'tortoise \'）到（NULL，\“defender \”，\“tortoise \”）看起来OKAY在windows中却没有在LINUX –

试试这个

php -r ' $mysqli = new mysqli("localhost","root","","monster"); 

    /* check connection */ 
    if (mysqli_connect_errno()) { 
     printf("Connect failed: %s\n", mysqli_connect_error()); 
     exit(); 
    } 

    /* return name of current default database */ 
    if ($result = $mysqli->query("INSERT INTO `earth` (`monster_id`, `type`, `name`) VALUES 
    (NULL, \'defender\', \'tortoise\') 
    ")) { 
     $row = $result->fetch_row(); 
     printf("Default database is %s.\n", $row[0]); 
     $result->close(); 
    } 
    $mysqli->close();'

来源

2014-01-29 09:22:07 NLSaini

仍然不能正常工作__ –

更改） – NLSaini

仍然不起作用NLSaini谢谢你的回复 –

我想你有一种语法错误那里。你正在用''开始你的php代码，并且在INSERT INTO语句中你也用'包装你的值'。也许这就是问题所在？

来源

2014-01-29 09:20:45 sobyte

试试这个

php -r ' $mysqli = new mysqli("localhost","root","","monster"); 

/* check connection */ 
if (mysqli_connect_errno()) { 
    printf("Connect failed: %s\n", mysqli_connect_error()); 
    exit(); 
} 
$type="defender"; 
$name="tortoise"; 
/* return name of current default database */ 
if ($result = $mysqli->query("INSERT INTO `earth` (`monster_id`, `type`, `name`) VALUES 
(NULL, $type, $name)")) 
{ 
    $row = $result->fetch_row(); 
    printf("Default database is %s.\n", $row[0]); 
    $result->close(); 
} 
$mysqli->close();'

来源

2014-01-29 09:30:20 Hamza

我已经让值变成了可变但仍然不起作用 –

PHP代码在PHP不行CLI

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