About the objects in the mysqli Extension Library $ mysqli_stmt $ mysqli-& gt; prepare ($ SQL); the prepare above belongs to the mysqli_stmt class? Why directly use $ mysqli-& gt; prepare? There is no relationship between the two. in this case, $ mysqli_stmlnewmysqli_stmt (); $ stmt $ mysql_stmt-& gt; pr issues with objects in the mysqli Extension Library.
$ Mysqli_stmt = $ mysqli-> prepare ($ SQL );
The prepare above belongs to the mysqli_stmt class? Why directly use $ mysqli-> prepare? It doesn't matter.
This should be the case.
$ Mysqli_stml = new mysqli_stmt ();
$ Stmt = $ mysql_stmt-> prepare ();
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Mysqli_stmt mysqli: prepare (string $ query)
Can you find the definition in the manual?
It indicates that prepare belongs to the mysqli class, the parameter is of the string type, and the returned value is an instance of the mysqli_stmt class.
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Discussion
I still don't quite understand .. Directly using this sentence $ stmt = $ mysqli-> prepare ($ SQL); is equivalent to creating an object $ stmt? But you should use new to create an object .. I don't understand this. I just typed it wrong ..
