Image 在Oracle中插入十六进制字符串作为图像

我想在Oracle10g中用十六进制字符串（图像的十六进制表示）更新包含图像的blob列。不幸的是，我唯一可以使用的是一个简单的查询，没有过程或其他任何东西。。。 我一直在尝试这个查询，但没有成功（PLSQL说它不是一个图像）：

当以这种方式使用时，它可以工作，但pl/sql既不能将其识别为映像，也不能识别应该从中读取的程序。我猜这是某种类型的类型转换问题，但我对OrSQL类型和转换不太熟悉

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任何建议都将不胜感激

你的方法是正确的。可能是输入十六进制错误？你怎么知道的

例如：

SQL> create table img (img blob);

Table created.

SQL> insert into img values (hextoraw('47494638396120002000f700000000000000330000660000990000cc0000ff002b00002b33002b66002b99002bcc002bff0055000055330055660055990055cc0055ff008
0000080330080660080990080cc0080ff00aa0000aa3300aa6600aa9900aacc00aaff00d50000d53300d56600d59900d5cc00d5ff00ff0000ff3300ff6600ff9900ffcc00ffff3300003300333300663300993300cc3300f
f332b00332b33332b66332b99332bcc332bff3355003355333355663355993355cc3355ff3380003380333380663380993380cc3380ff33aa0033aa3333aa6633aa9933aacc33aaff33d50033d53333d56633d59933d5cc3
3d5ff33ff0033ff3333ff6633ff9933ffcc33ffff6600006600336600666600996600cc6600ff662b00662b33662b66662b99662bcc662bff6655006655336655666655996655cc6655ff668000668033668066668099668
0cc6680ff66aa0066aa3366aa6666aa9966aacc66aaff66d50066d53366d56666d59966d5cc66d5ff66ff0066ff3366ff6666ff9966ffcc66ffff9900009900339900669900999900cc9900ff992b00992b33992b66992b9
9992bcc992bff9955009955339955669955999955cc9955ff9980009980339980669980999980cc9980ff99aa0099aa3399aa6699aa9999aacc99aaff99d50099d53399d56699d59999d5cc99d5ff99ff0099ff3399ff669
9ff9999ffcc99ffffcc0000cc0033cc0066cc0099cc00cccc00ffcc2b00cc2b33cc2b66cc2b99cc2bcccc2bffcc5500cc5533cc5566cc5599cc55cccc55ffcc8000cc8033cc8066cc8099cc80cccc80ffccaa00ccaa33cca
a66ccaa99ccaaccccaaffccd500ccd533ccd566ccd599ccd5ccccd5ffccff00ccff33ccff66ccff99ccffccccffffff0000ff0033ff0066ff0099ff00ccff00ffff2b00ff2b33ff2b66ff2b99ff2bccff2bffff5500ff553
3ff5566ff5599ff55ccff55ffff8000ff8033ff8066ff8099ff80ccff80ffffaa00ffaa33ffaa66ffaa99ffaaccffaaffffd500ffd533ffd566ffd599ffd5ccffd5ffffff00ffff33ffff66ffff99ffffccffffff0000000
0000000000000000021f904010000fc002c00000000200020000008ff00371429b2619fc183080dbe899630e140810407366c2870e1c47d103764bc887020135017230e8cc8f1601168251f1661387119ca7dd194edabb3a
c24cb92758c286482b188c5921ca31154267483b29120815e7c03d10e44262f955e84080ad4c88755056e80131528c1a4a0b41601ab72a5d20d490d5a458b30ec489940d31eacda10d4929a0da325d5bb4fae5abf7db94e7
c53446161b673bfb68d28f8a0d0224405421b4896aa4168102923842370099c8146b4561d29da6d91253783d68908ca944881a309323dcd59e24168b535f7150b566051a204bb46232cd0cee2b18bf77d36b87a22b437804
1194f08121a5c972d3f3634155d296ecb8991275156ba6cb568b51111bb2e28f539d5b5a46317967a713893919f05d2cf4b78a5e468beed979032f715449861023a07074c7b016692562595a7547a22a5765b6d6791f6d04
5e51134a158247134d984230504003b'));

1 row created.

SQL> commit;

Commit complete.

然后在我的PL/SQLIDE（PL/SQLDeveloper）中，我看到。。

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我通过使用十六进制编辑器打开.jpg文件来获得十六进制字符串。可以很好地使用MSSQL。可能是更新操作失败了吗？因为我不习惯插入新行，只是更新现有行。@BorislavGeorgiev update vs insert应该没关系。您的文本编辑器不会翻转字节，是吗？例如，我的意思是gif以47 49 46 38开头，但有时程序会将其显示为4947 3846等。@BorislavGeorgiev p.s.如果您运行我提供的测试用例，您看到图像了吗？（作为insert+as更新）嘿，非常感谢您的回复…似乎我得到了错误的数据类型（hexstring作为FFD8FF000104A464946000101010048000480000FFDB），因为它最初在MSSQL中是这样工作的。使用您提供的数据时，它就像一个符咒！非常感谢！！！

SQL> create table img (img blob);

Table created.

SQL> insert into img values (hextoraw('47494638396120002000f700000000000000330000660000990000cc0000ff002b00002b33002b66002b99002bcc002bff0055000055330055660055990055cc0055ff008
0000080330080660080990080cc0080ff00aa0000aa3300aa6600aa9900aacc00aaff00d50000d53300d56600d59900d5cc00d5ff00ff0000ff3300ff6600ff9900ffcc00ffff3300003300333300663300993300cc3300f
f332b00332b33332b66332b99332bcc332bff3355003355333355663355993355cc3355ff3380003380333380663380993380cc3380ff33aa0033aa3333aa6633aa9933aacc33aaff33d50033d53333d56633d59933d5cc3
3d5ff33ff0033ff3333ff6633ff9933ffcc33ffff6600006600336600666600996600cc6600ff662b00662b33662b66662b99662bcc662bff6655006655336655666655996655cc6655ff668000668033668066668099668
0cc6680ff66aa0066aa3366aa6666aa9966aacc66aaff66d50066d53366d56666d59966d5cc66d5ff66ff0066ff3366ff6666ff9966ffcc66ffff9900009900339900669900999900cc9900ff992b00992b33992b66992b9
9992bcc992bff9955009955339955669955999955cc9955ff9980009980339980669980999980cc9980ff99aa0099aa3399aa6699aa9999aacc99aaff99d50099d53399d56699d59999d5cc99d5ff99ff0099ff3399ff669
9ff9999ffcc99ffffcc0000cc0033cc0066cc0099cc00cccc00ffcc2b00cc2b33cc2b66cc2b99cc2bcccc2bffcc5500cc5533cc5566cc5599cc55cccc55ffcc8000cc8033cc8066cc8099cc80cccc80ffccaa00ccaa33cca
a66ccaa99ccaaccccaaffccd500ccd533ccd566ccd599ccd5ccccd5ffccff00ccff33ccff66ccff99ccffccccffffff0000ff0033ff0066ff0099ff00ccff00ffff2b00ff2b33ff2b66ff2b99ff2bccff2bffff5500ff553
3ff5566ff5599ff55ccff55ffff8000ff8033ff8066ff8099ff80ccff80ffffaa00ffaa33ffaa66ffaa99ffaaccffaaffffd500ffd533ffd566ffd599ffd5ccffd5ffffff00ffff33ffff66ffff99ffffccffffff0000000
0000000000000000021f904010000fc002c00000000200020000008ff00371429b2619fc183080dbe899630e140810407366c2870e1c47d103764bc887020135017230e8cc8f1601168251f1661387119ca7dd194edabb3a
c24cb92758c286482b188c5921ca31154267483b29120815e7c03d10e44262f955e84080ad4c88755056e80131528c1a4a0b41601ab72a5d20d490d5a458b30ec489940d31eacda10d4929a0da325d5bb4fae5abf7db94e7
c53446161b673bfb68d28f8a0d0224405421b4896aa4168102923842370099c8146b4561d29da6d91253783d68908ca944881a309323dcd59e24168b535f7150b566051a204bb46232cd0cee2b18bf77d36b87a22b437804
1194f08121a5c972d3f3634155d296ecb8991275156ba6cb568b51111bb2e28f539d5b5a46317967a713893919f05d2cf4b78a5e468beed979032f715449861023a07074c7b016692562595a7547a22a5765b6d6791f6d04
5e51134a158247134d984230504003b'));

1 row created.

SQL> commit;

Commit complete.