PHP Parsing/Syntax Error Unexpected End of File

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PHP syntax error “unexpected $end” 7 answers

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You have not clased the mysql_connect errorno() brackets. Use the code below

<?php

// Create connection
$con=mysqli_connect("host","username","password","db");

// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// escape variables for security
 $name = mysqli_real_escape_string($con, $_POST['name']);

 $sql="INSERT INTO testing (Name)
 VALUES ('$name')";

 if (!mysqli_query($con,$sql)) {
  die('Error: ' . mysqli_error($con));
}

 echo "1 record added";

 mysqli_close($con);
?>

Hope this helps you

---

你还没有关闭你的代码

<?php

// Create connection
$con=mysqli_connect("host","username","password","db");

// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
die; // append this
 } //apend this 
// escape variables for security
 $name = mysqli_real_escape_string($con, $_POST['name']);

 $sql="INSERT INTO testing (Name)
 VALUES ('$name')";

 if (!mysqli_query($con,$sql)) {
  die('Error: ' . mysqli_error($con));
}

 echo "1 record added";

 mysqli_close($con);


?>

---

你的if (mysqli_connect_errno()) {没有关闭

<?php

// Create connection
$con=mysqli_connect("host","username","password","db");

// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// escape variables for security
 $name = mysqli_real_escape_string($con, $_POST['name']);

 $sql="INSERT INTO testing (Name)
 VALUES ('$name')";

 if (!mysqli_query($con,$sql)) {
  die('Error: ' . mysqli_error($con));
}

 echo "1 record added";

 mysqli_close($con);
?>

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