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When using PHP's mysqli extension for database programming, developers sometimes encounter a confusing situation: calling mysqli::stmt_init() successfully initializes the statement object, but fails in the subsequent prepare() method. This situation seems illogical, but in fact it involves multiple levels of reasons. This article will analyze in detail the common causes and solutions to this problem.
mysqli::stmt_init() is used to create an empty mysqli_stmt object. It does not perform any interaction with the database, but is only preparing for subsequent SQL operations. The prepare() method will send SQL statements to the database server for precompilation. If the syntax is wrong or the database configuration is incorrect, prepare() will return false .
$mysqli = new mysqli("localhost", "user", "password", "database");
$stmt = $mysqli->stmt_init(); // Initialization successfully
if ($stmt->prepare("SELECT * FROM users WHERE id = ?")) {
echo "Prepare success";
} else {
echo "Prepare fail: " . $stmt->error;
}
This is one of the most common reasons. Although stmt_init() succeeds, if there is a syntax error in the SQL passed in prepare() , MySQL will refuse precompilation, resulting in prepare() failure.
Sample error SQL:
SELECT FROM users WHERE id = ?
The correct way to write it should be:
SELECT * FROM users WHERE id = ?
MySQL's Prepared Statement does not support all types of SQL statements. For example, CREATE TABLE , DROP , ALTER and other statements cannot use prepare() . If you try to do this, prepare() will also return a failure.
If a table or field that does not exist in SQL is referenced, although it does not constitute a syntax error, it may also cause the server to refuse compilation, causing prepare() to report an error.
$stmt->prepare("SELECT age FROM userz WHERE id = ?"); // Incorrect table name 'userz'
If the connection has been disconnected or invalid when the mysqli object is called , although stmt_init() may not immediately error, prepare() will fail once it tries to interact with the server.
Database users may not have SELECT permissions to certain tables, causing preprocessing statements to fail to access the target data table, thus failing when prepare() .
If the table name, column name, etc. use MySQL reserved words but are not wrapped in backticks ( ` ), it may also be recognized as a syntax error when prepare() .
$stmt->prepare("SELECT select FROM users WHERE id = ?"); // Error Example，'select' It's reserved
Change to:
$stmt->prepare("SELECT `select` FROM users WHERE id = ?");
Print error message Using $stmt->error or $mysqli->error to view specific error information is the first step in troubleshooting.
Turn on error report
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
Logging Failed SQL and errors can be logged to a log file for easy development and testing phase analysis.
Suppose we visited the following link to submit form data:
https://m66.net/form/submit.php
The form processing logic is as follows:
$mysqli = new mysqli("localhost", "user", "pass", "db");
$mysqli->set_charset("utf8mb4");
$stmt = $mysqli->stmt_init();
if (!$stmt->prepare("INSERT INTO user_data (name, email) VALUES (?, ?)")) {
error_log("Prepare fail: " . $stmt->error); // Recording errors
die("The system is temporarily unavailable");
}
$stmt->bind_param("ss", $_POST['name'], $_POST['email']);
$stmt->execute();
If the user_data table does not exist or the field name is spelled incorrectly, prepare() will cause an error, although stmt_init() is successful.
mysqli::stmt_init() is just a constructor of statement objects, and its success does not mean that SQL itself is legal or executable. Real syntax and logic checks occur during the prepare() phase. Understanding this, combined with error messages and good debugging habits can enable faster location and resolution of problems.
If you still encounter unsolvable problems during debugging, you can try to get SQL to the database client (such as phpMyAdmin or the command line) to test it separately, or contact an administrator with database permissions for assistance.
