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Full Discussion: Printf statement
Top Forums Shell Programming and Scripting Printf statement Post 302501057 by jim mcnamara on Wednesday 2nd of March 2011 10:44:19 AM
Old 03-02-2011
When printf encounters a character that is not [:digit:] or a dot or a - you get that error. Nothing else does that. The data is the problem, not necessarily the code.

It can be an unprintable character. try
Code:
echo $4 | od

And you are positive that the 000 characters are not character O rather than numeric 0
 

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MYSQLI_STMT_ERROR(3)							 1						      MYSQLI_STMT_ERROR(3)

mysqli_stmt::$error - Returns a string description for last statement error

       Object oriented style

SYNOPSIS
string$mysqli_stmt->error () DESCRIPTION
Procedural style string mysqli_stmt_error (mysqli_stmt $stmt) Returns a string containing the error message for the most recently invoked statement function that can succeed or fail. PARAMETERS
o $ stmt -Procedural style only: A statement identifier returned by mysqli_stmt_init(3). RETURN VALUES
A string that describes the error. An empty string if no error occurred. EXAMPLES
Example #1 Object oriented style <?php /* Open a connection */ $mysqli = new mysqli("localhost", "my_user", "my_password", "world"); /* check connection */ if (mysqli_connect_errno()) { printf("Connect failed: %s ", mysqli_connect_error()); exit(); } $mysqli->query("CREATE TABLE myCountry LIKE Country"); $mysqli->query("INSERT INTO myCountry SELECT * FROM Country"); $query = "SELECT Name, Code FROM myCountry ORDER BY Name"; if ($stmt = $mysqli->prepare($query)) { /* drop table */ $mysqli->query("DROP TABLE myCountry"); /* execute query */ $stmt->execute(); printf("Error: %s. ", $stmt->error); /* close statement */ $stmt->close(); } /* close connection */ $mysqli->close(); ?> Example #2 Procedural style <?php /* Open a connection */ $link = mysqli_connect("localhost", "my_user", "my_password", "world"); /* check connection */ if (mysqli_connect_errno()) { printf("Connect failed: %s ", mysqli_connect_error()); exit(); } mysqli_query($link, "CREATE TABLE myCountry LIKE Country"); mysqli_query($link, "INSERT INTO myCountry SELECT * FROM Country"); $query = "SELECT Name, Code FROM myCountry ORDER BY Name"; if ($stmt = mysqli_prepare($link, $query)) { /* drop table */ mysqli_query($link, "DROP TABLE myCountry"); /* execute query */ mysqli_stmt_execute($stmt); printf("Error: %s. ", mysqli_stmt_error($stmt)); /* close statement */ mysqli_stmt_close($stmt); } /* close connection */ mysqli_close($link); ?> The above examples will output: Error: Table 'world.myCountry' doesn't exist. SEE ALSO
mysqli_stmt_errno(3), mysqli_stmt_sqlstate(3). PHP Documentation Group MYSQLI_STMT_ERROR(3)
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