我正在尝试制作一个html表单，填充后将插入到我的sql数据库中。但是我得到这个错误，没有选择数据库[重复]

问题描述：

This question already has an answer here:

Can I mix MySQL APIs in PHP? 4 answers

I get the error

no database selected

I am using wamp. I am very new to php and sql. here is the php html. the inputs comes from a html file first when the submit button is pressed.

<?php

$link = mysqli_connect("localhost", "erthiph", "");

if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
    $db_select = mysql_select_db( 'inewsfeed' ,$connection);
    if (!$db_select) {
        die("Database selection failed:: " . mysql_error());
        }
}

$Qid = mysqli_real_escape_string($link, $_POST['Qid']);
$Mclass = mysqli_real_escape_string($link, $_POST['Mclass']);
$Sclass = mysqli_real_escape_string($link, $_POST['Sclass']);
$Question = mysqli_real_escape_string($link, $_POST['Question']);
$Answer = mysqli_real_escape_string($link, $_POST['Answer']);
$Doc = mysqli_real_escape_string($link, $_POST['Doc']);
$Time = mysqli_real_escape_string($link, $_POST['Time']);

$sql = "INSERT INTO feed (Qid, Mclass, Sclass, Question, Answer, Doc, Time) VALUES ('$Qid', '$Mclass', '$Sclass', '$Question', '$Answer', '$Doc', '$Time')";
if(mysqli_query($link, $sql)){
    echo "Records added successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

mysqli_close($link);
?>

</div>

此问题已经存在这里有一个答案： p>

我可以在PHP中混合MySQL API吗？ 4 answers span > li> ul> div>

我收到错误 p>

没有选择数据库 p> blockquote>

我正在使用wamp。我是php和sql的新手。这是php html。当按下提交按钮时，输入首先来自html文件。 p>

 ＆lt;？php 
 
 $ link = mysqli_connect（“localhost”，“erthiph”，  “”）; 
 
if（$ link === false）{
 die（“ERROR：无法连接。”。mysqli_connect_error（））; 
 $ db_select = mysql_select_db（'inewsfeed'，$ connection）;  
 if（！$ db_select）{
 die（“数据库选择失败::”。mysql_error（））; 
} 
} 
 
 $ Qid = mysqli_real_escape_string（$ link，$ _POST ['Qid'  ]）; 
 $ Mclass = mysqli_real_escape_string（$ link，$ _POST ['Mclass']）; 
 $ Sclass = mysqli_real_escape_string（$ link，$ _POST ['Sclass']）; 
 $ Question = mysqli_real_escape_string（$ link  ，$ _POST ['Question']）; 
 $ Answer = mysqli_real_escape_string（$ link，$ _POST ['Answer']）; 
 $ Doc = mysqli_real_escape_string（$ link，$ _POST ['Doc']）; 
  $ Time = mysqli_real_escape_string（$ link，$ _POST ['Time']）; 
 
 $ sql =“INSERT INTO feed（Qid，Mclass，Sclass，Question，Answer，Doc，Time）VALUES（'$ Qid'，  '$ Mclass'，'$ Sclass'，'$ Question'，'$ Answer'，'$ Doc'，'$ Time'）“; 
  if（mysqli_query（$ link，$ sql））{
 echo“记录添加成功。”; 
} else {
 echo“错误：无法执行$ sql。  “.mysqli_error（$ link）; 
} 
 
mysqli_close（$ link）; 
？＆gt; 
  code>  pre> 
  div>

答

What @saty said is correct.You can try like this also,

mysqli_connect("localhost","username","password","dbname");

答

Because you are mixing mysql add mysqli

Instead of this

mysql_select_db( 'inewsfeed' ,$connection);

use

mysqli_select_db($link,"inewsfeed");

Read http://php.net/manual/en/mysqli.select-db.php

我正在尝试制作一个html表单，填充后将插入到我的sql数据库中。 但是我得到这个错误，没有选择数据库[重复]

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我正在尝试制作一个html表单，填充后将插入到我的sql数据库中。但是我得到这个错误，没有选择数据库[重复]