PHP syntax error “unexpected $end”

php syntax syntax-error

43,600

Solution 1

$result = mysql_query($sql, $connection) or die(mysql_error());
if ($result) {
    $msg = "<p>" .$_POST[table_name]." has been created!</p>";
}

you missing a } in your last if statement, and your for loop is missing a } too

for ($i = 0; $i < count($_POST[field_name]); $i++) {
    $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
    if ($_POST[field_length][$i] !="") {
      $sql .=" (".$_POST[field_length][$i]."),";
    } else {
        $sql .=",";
    }
}

Solution 2

This error message means that a control structure block isn’t closed properly. In your case the closing } of some of your control structures like the for loop or the last if are missing.

You should use proper indentation and an editor that highlights bracket pairs to have a visual aid to avoid such errors.

Solution 3

in your php.ini (php configuration) change :

short_open_tag = Off

you opened php tag shortly at line 1
just find and replace all <? with <?php

Solution 4

You must close the for expression block:

for ($i = 0; $i < count($_POST[field_name]); $i++) {
    $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
    // ...
}

Solution 5

Your for loop is not terminated. You are missing a }

for ($i = 0; $i < count($_POST[field_name]); $i++) { $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i]; }

And as pointed by others there is also a missing } for the last if statement:

if ($result) { $msg = "< p>" .$_POST[table_name]." has been created!< /p>"; }

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43,600

Author by

Jacksta

Updated on July 05, 2022

Comments

Jacksta almost 2 years

I have 3 files 1) show_createtable.html 2) do_showfielddef.php 3) do_showtble.php

1) First file is for creating a new table for a data base, it is a fom with 2 inputs, Table Name and Number of Fields. THIS WORKS FINE!

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<h1>Step 1: Name and Number</h1>
<form method="post" action="do_showfielddef.php" />
<p><strong>Table Name:</strong><br />
<input type="text" name="table_name" size="30" /></p>
<p><strong>Number of fields:</strong><br />
<input type="text" name="num_fields" size="30" /></p>
<p><input type="submit" name="submit" value="go to step2" /></p>
</form>


</body>
</html>

2) this script validates fields and createa another form to enter all the table rows. This for also WORKS FINE!

<?php
//validate important input
if ((!$_POST[table_name]) || (!$_POST[num_fields])) {
    header( "location: show_createtable.html");
           exit;
}

//begin creating form for display
$form_block = "
<form action=\"do_createtable.php\" method=\"post\">
<input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\">
<table cellspacing=\"5\" cellpadding=\"5\">
  <tr>
    <th>Field Name</th><th>Field Type</th><th>Table Length</th>
  </tr>";

//count from 0 until you reach the number fo fields
for ($i = 0; $i <$_POST[num_fields]; $i++) {
  $form_block .="
  <tr>
  <td align=center><input type=\"texr\" name=\"field name[]\"
  size=\"30\"></td>
  <td align=center>
    <select name=\"field_type[]\">
        <option value=\"char\">char</option>
        <option value=\"date\">date</option>
        <option value=\"float\">float</option>
        <option value=\"int\">int</option>
        <option value=\"text\">text</option>
        <option value=\"varchar\">varchar</option>
        </select>
  </td>
  <td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\">
  </td>
</tr>";
}

//finish up the form 
$form_block .= "
<tr>
    <td align=center colspan=3><input type =\"submit\" value=\"create table\">
    </td>
</tr>
</table>
</form>";

?>

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Create a database table: Step 2</title>
</head>

<body>
<h1>defnie fields for <? echo "$_POST[table_name]"; ?> 
</h1>
<? echo "$form_block"; ?>

</body>
</html>

Problem is here 3) this form creates the tables and enteres them into the database. I am getting an error on line 37 "Parse error: syntax error, unexpected $end in /home/admin/domains/domaina.com.au/public_html/do_createtable.php on line 37"

<?
$db_name = "testDB";

$connection = @mysql_connect("localhost", "admin_user", "pass")
    or die(mysql_error());

$db = @mysql_select_db($db_name, $connection)
    or die(mysql_error());

$sql = "CREATE TABLE $_POST[table_name](";
    for ($i = 0; $i < count($_POST[field_name]); $i++) {
        $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
    if ($_POST[field_length][$i] !="") {
        $sql .=" (".$_POST[field_length][$i]."),";
        } else {
            $sql .=",";
        }
$sql = substr($sql, 0, -1);
$sql .= ")";

$result = mysql_query($sql, $connection) or die(mysql_error());
if ($result) {
    $msg = "<p>" .$_POST[table_name]." has been created!</p>";

?>

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Create A Database Table: Step 3</title>
</head>

<body>
<h1>Adding table to <? echo "$db_name"; ?>...</h1>
<? echo "$msg"; ?>
</body>
</html>