PHP Warning: PDOStatement::execute(): SQLSTATE[HY093]: Invalid parameter number: no parameters were bound

Question

I use PDO to execute sql

"INSERT INTO zhushou_cost_uid
(uid,imei,wmac,imsi,channel,supplier,uuid,brand,device_model,os,os_version,app_version,promotion_method,log_source,takeup_date)
 VALUES
('863207010118070','863207010118070','02037ff459cb','460025323359694','sc-hjcx_ins_cgq','','�ܟ*c�1�]�y�.���#���h���!�o ��z�!Y�~��t8�KOd�xd]���sm����n%$����H����[?�p���M����','KINGSUN','KINGSUN S6','Android','4.1.2','3.2','','1','2015-11-29 03:21:21')",

PHP code:

$db = $this->getWritableDB();
$stmt = $db->prepare($sql);
$exec = $stmt->execute();

the data of uuid is dirty data, and in our log it is

"uuid":"�ܟ*c�1�]�y�.��#��\u0015�h\u001a��!\u001c�\u0013o �\u0013�z�\u0000!Y�~��t8�KOd�xd]�\u0001��sm\u0016��\u0001n\u0013%$��H��[\u0003?�p��M��\u001a��"

I got the result

SQLSTATE[HY093]: Invalid parameter number: no parameters were bound.

When I try to copy the sql onto the terminator, exiting the mysql login status. I think there is something wrong in uuid. But I can not figure out it. Can anyone help me? Thank you very much!

Copy the SQL to the terminator which leads to exit mysql login status, I am so confused. — wlstony
– wlstony, Commented Dec 8, 2015 at 8:30
Please post your php code. Possible duplicate of stackoverflow.com/questions/20436745/… — maxhb
– maxhb, Commented Dec 8, 2015 at 8:35
Our php code likes this(the SQL is in my question): $db = $this->getWritableDB(); $stmt = $db->prepare($sql); $exec = $stmt->execute(); — wlstony
– wlstony, Commented Dec 8, 2015 at 8:48
Update your question with that code, don't leave it in a comment. — deceze
– deceze ♦, Commented Dec 8, 2015 at 8:50

deceze · Accepted Answer · 2015-12-08 08:53:44Z

5

$stmt = $db->prepare($sql);

Since you're preparing your statement, if there's anything in it that can be interpreted as a question mark or colon, it will be taken as a placeholder and you're expected to then pass values for it in the execute step. Since this is not actually what you're intending, don't prepare the statement if you don't intend to have placeholders in it. Instead:

$db->exec($sql);

Having said that, it's suspicious that you're passing a fully formed SQL query in $sql; perhaps you should be rewriting this whole thing so you do have actual placeholders in your query and are passing the actual values separately to execute.

answered Dec 8, 2015 at 8:53

deceze♦

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1 Comment

wlstony Over a year ago

I tried what you said, the SQL executed successfully, thanks very much.

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