从mysqli_query获取成功/失败响应 [英] get success/failure response from mysqli_query

问题描述

我正在从用户那里收集信息，然后将它们添加到表中.

I'm gathering info from a user and then adding them to a table.

$insert = "INSERT INTO jos_activeagents (RINGPHONE, AGENTUID, FNAME, LNAME) VALUES ('[phone]','".$result['AGTBRDIDMM']."','".$result['AGTFNAME']."','".$result['AGTLNAME']."')";

$set = mysqli_query($link,$insert);

AGENTUID是唯一键.如果用户尝试使用重复的唯一密钥提交，我会收到一条错误消息(当然).

AGENTUID is a unique key. If a user tries to submit with a duplicate unique key, I get an error (of course).

现在，我该如何知道是否以及何时发生错误，然后将响应返回到页面呢?我知道mysqli_get_warnings()，但是PHP手册没有显示任何示例.

Now, how would I go about knowing if and when an error occurred and then putting a response back to the page? I know of mysqli_get_warnings(), but the PHP manual doesn't show any examples.

我也尝试过先在表中查找AGENTUID:

I have also tried looking for the AGENTUID in the table first:

$check = "SELECT * FROM jos_activeagents WHERE AGENTUID = '".$agt."'";

$runcheck = mysqli_query($link,$check);

$rescheck = mysqli_fetch_assoc($runcheck);

if($rescheck != null){

    echo 'This Agent ID is already enrolled.'

}

但这似乎草率.有更好的方法吗?

But this seems sloppy. Is there a better way do this?

推荐答案

您可以使用 mysqli_error() 查看是否发生错误

You can use mysqli_error() to see if an error occurred

if (mysqli_error($runcheck ))
{
   // an error eoccurred
}

在您的特定示例中，最好在插入之前 检查行是否存在.您的示例很接近，但最好使用 mysqli_num_rows() :

In your particular example you're better of checking if the row exists before doing the insert. Your example is close but would better using mysqli_num_rows():

$check = "SELECT * FROM jos_activeagents WHERE AGENTUID = '".$agt."'";
$runcheck = mysqli_query($link,$check);
if (mysqli_num_rows($runcheck) > 0)
{
    // username in use
}

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