[spc_123]

No, I did not knowingly build any constraints. In fact, if I use DBManager instead of SQLyog, I don't have any problem creating the relationships. In fact, that was the only way I managed to create the foreign key section displayed in the following table. I'd prefer to use SQLyog if possible. Here is the first table info:

Column Information For – cyberdb.cs_component

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Field Type Null Key Default Extra

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---

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id smallint(8) PRI (NULL) auto_increment

component_type_id smallint(8) YES MUL (NULL)

segment_id smallint(8) YES (NULL)

comp_conf_id smallint(8) YES (NULL)

Index Information For – cyberdb.cs_component

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Table Non_unique Key_name Seq_in_index Column_name Collation Cardinality Sub_part Packed Null Index_type Comment

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---

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---

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---

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cs_component 0 PRIMARY 1 id A 0 (NULL) (NULL) BTREE

cs_component 1 component_type_id 1 component_type_id A 0 (NULL) (NULL) YES BTREE

DDL Information For – cyberdb.cs_component

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Table Create Table

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cs_component CREATE TABLE `cs_component` (

`id` smallint(8) NOT NULL auto_increment,

`component_type_id` smallint(8) default NULL,

`segment_id` smallint(8) default NULL,

`comp_conf_id` smallint(8) default NULL,

PRIMARY KEY (`id`),

KEY `component_type_id` (`component_type_id`),

FOREIGN KEY (`component_type_id`) REFERENCES `cs_component_type` (`id`)

TYPE=InnoDB

*************************************************

here is the second:

Column Information For – cyberdb.cs_component_type

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Field Type Null Key Default Extra

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---

---

---

---

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id smallint(8) PRI (NULL) auto_increment

name varchar(20) YES (NULL)

tier_name smallint(2) YES (NULL)

Index Information For – cyberdb.cs_component_type

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Table Non_unique Key_name Seq_in_index Column_name Collation Cardinality Sub_part Packed Null Index_type Comment

---

---

---

---

---

---

---

---

---

---

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cs_component_type 0 PRIMARY 1 id A 13 (NULL) (NULL) BTREE

DDL Information For – cyberdb.cs_component_type

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Table Create Table

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cs_component_type CREATE TABLE `cs_component_type` (

`id` smallint(8) NOT NULL auto_increment,

`name` varchar(20) default NULL,

`tier_name` smallint(2) default NULL,

PRIMARY KEY (`id`)

TYPE=InnoDB

*************************************************************

Any ideas?

[spc_123]

Then, right after I get this message, I continually get the Error 1100 again…..which is then resolved by disconnecting and reconnecting again. I resolved the problem by dropping the problem tables and recreating them.

[spc_123]

I disconnected/reconnected, and it seems that I can open my tables again…however, when I try to delete a relationship on a form I get: Error No. 1064 You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '?, modified, im_id from `cyberdb`.`cs_operator`' at line 1

I'm doing this via SQLyog, and not personally running any SQL

[spc_123]

Resolved this issue….I had several indexes on this table. Once I deleted the indexes, I was able to delete the relationships…..Is this as designed? Thanks for being there anyway…

[spc_123]

Thank you both! As it turns out, I was running v.3 not 4 as I thought, and hadn't run the mysql-max.exe. I removed it and installed v4, made the modifications to the my.ini manually, removed the service, and ran the mysql-max.exe. Seems to be fine! Thanks again, you guys saved my posterior.

[spc_123]

THANK YOU RiTesh!!! Less filling Works great!

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